#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);


/*
 * Complete the 'closestNumbers' function below.
 *
 * The function accepts INTEGER_ARRAY numbers as parameter.
 */

void closestNumbers(vector<int> numbers)
{
    int n = numbers.size();

    if (n <= 1) return;

    sort(numbers.begin(), numbers.end());

    int minDiff = numbers[1] - numbers[0];

    for (int i = 2; i < n; ++i)
    {
        minDiff = min(minDiff, numbers[i] - numbers[i - 1]);
    }

    unordered_map<int, int> m;

    for (int i = 1; i < n; ++i)
    {
        if (numbers[i] - numbers[i - 1] == minDiff)
        {
            printf("%d %d\n", numbers[i - 1], numbers[i]);
        }
    }
}
int main()
{
    string numbers_count_temp;
    getline(cin, numbers_count_temp);

    int numbers_count = stoi(ltrim(rtrim(numbers_count_temp)));

    vector<int> numbers(numbers_count);

    for (int i = 0; i < numbers_count; i++) {
        string numbers_item_temp;
        getline(cin, numbers_item_temp);

        int numbers_item = stoi(ltrim(rtrim(numbers_item_temp)));

        numbers[i] = numbers_item;
    }

    closestNumbers(numbers);

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

#include <bits/stdc++.h>

using namespace std;

vector<int> solution(vector<int> &A, int K)
{
    int n = A.size();
    if (n == 0) return A;

    reverse(A.begin(), A.end());

    K %= n;

    reverse(A.begin(), A.begin() + K);
    reverse(A.begin() + K, A.end());
    return A;
}

class Solution {
public:
    int countSquares(int m, int n) {
        int count = 0;
        for (int i = 1; i <= min(m, n); ++i) {
            count += (m - i + 1) * (n - i + 1);
        }
        return count;
    }
};

#include <bits/stdc++.h>

using namespace std;

int solution(int X, int Y, int D)
{
    return ceil((Y - X) * 1.0 / D);
}

#include <bits/stdc++.h>

using namespace std;

int solution(int N)
{
    int maxGap = 0, curGap = 0;

    // 這個 loop 移除 N 前面的所有 0，直到遇到第一個 1 為止。這是因為 leading zero 不會形成有效的間隙。
    while (N > 0 && N % 2 == 0)
    {
         N /= 2;
    }

    // 走訪剩餘的二進位數字
    while (N > 0)
    {
        int digits = N % 2;

        // digits 等於 0 代表是在 gap 中
        if (digits == 0)
        {
            ++curGap;
        }
        else
        {
            // digits 不等於 0 代表 gap 結束
            if (curGap)
            {
                // 這時候計算 maxGap 最大是多少，並且 reset curGap
                maxGap = max(curGap, maxGap);
                curGap = 0;
            }

        }
        N /= 2;
    }
    return maxGap;
}

#include <bits/stdc++.h>

using namespace std;

int solution(vector<int> &A)
{
    unordered_map<int, int> countMap;
    for (auto a : A) ++countMap[a];

    for (auto [k, v] : countMap)
    {
        if (v % 2 == 0) continue;
        return k;
    }
}

#include <bits/stdc++.h>
using namespace std;

int solution(vector<int> &A)
{
    int n = A.size();
    int head = A[0];
    int tail = accumulate(A.begin() + 1, A.end(), 0);

    int minDiff = abs(head - tail);

    for (int i = 1; i < n - 1; ++i)
    {
        head += A[i];
        tail -= A[i];
        int cur = abs(head - tail);
        minDiff = min(minDiff, cur);
    }
    return minDiff;
}

#include <bits/stdc++.h>

using namespace std;

int solution(int X, vector<int> &A)
{
    set<int> st;
    int n = A.size();
    for (int i = 0; i < n; ++i)
    {
        st.insert(A[i]);
        if (st.size() == X)
        {
            return i;
        }
    }
    return -1;
}

string solution(string& S)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

string solution(string &S)
{
    vector<int> seen;

    // 這邊先建立一個 seen 數列，方便操作
    // 例如以 a?bb 來說，就會 return [0, 0, 1, 0, -1, -1, 0, 0]
    for (char c : s)
    {
        if (c == 'a')
        {
            seen.push_back(1);
        }
        else if (c == 'b')
        {
            seen.push_back(-1);
        }
        else
        {
            seen.push_back(0);
        }
    }

    seen.insert(seen.begin(), {0, 0});
    seen.insert(seen.end(), {0, 0});

    int n = seen.size();

    auto helper = [&seen](int i)
    {
        // 如果遇到問號的話
        if (seen[i] == 0)
        {
            vector<int> sums = {seen[i - 2] + seen[i - 1], seen[i - 1] + seen[i + 1], seen[i + 1] + seen[i + 2]};
            if (find(sums.begin(), sums.end(), 2) != sums.end())
            {
                 // 如果相鄰的兩個 'a'，則填入 'b'
                seen[i] = -1;
            } else if (find(sums.begin(), sums.end(), -2) != sums.end())
            {
                // 如果相鄰的兩個 'b'，則填入 'a'
                seen[i] = 1;
            }
        }
    };

    // 由左到右走訪
    for (int i = 2; i < n - 2; ++i)
    {
        helper(i);
    }

    // 由右到左走訪
    for (int i = n - 3; i > 1; --i)
    {
        helper(i);
    }


    // 由左到右走訪，如果遇到 0
    for (int i = 2; i < n - 2; ++i) {
        if (seen[i] == 0)
        {
            // 檢查前一個元素是 1 還是 -1
            seen[i] = (seen[i - 1] > 0) ? -1 : 1;
        }
    }

    // 將數列填入 res
    string res;
    for (int i = 2; i < n - 2; ++i)
    {
        if (seen[i] == 1)
        {
            res += 'a';
        } else if (seen[i] == -1) {
            res += 'b';
        }
    }

    return res;
}

int main()
{
    vector<string> s = {"a?bb", "??abb", "a?b?aa", "aa??aa"};
    for (auto c : s)
    {
        cout << solution(s) << endl;
    }
}