//最小堆算法
/*
#include <iostream>
#include <queue>

using namespace std;

typedef long long ll;

const int N = 1e5 + 10;

struct node {
	ll sum;
	int i;
	int j;

	bool operator < (const node& x) const {
		return this->sum > x.sum;
	}
};

int a[N], b[N];
int n;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);

	cin >> n;

	for (int i = 1; i <= n; ++i) cin >> a[i];
	for (int i = 1; i <= n; ++i) cin >> b[i];

	priority_queue<node> heap;

	for (int i = 1; i <= n; ++i) {
		heap.push({ a[1] + b[i], 1, i});
	}

	for (int k = 1; k <= n; ++k) {
		node t = heap.top(); heap.pop();
		ll sum = t.sum;
		int i = t.i, j = t.j;
		cout << sum << " ";
		if(i + 1 <= n)
			heap.push({ (ll)a[i + 1] + b[j], i + 1, j });
	}
	return 0;
}
*/

//无敌的暴力枚举
/*
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

const int N = 1e5 + 10;

int arr1[N], arr2[N];
int n;

int main() {
    priority_queue<int> result;  // 最大堆，用来存储前 N 个最小的和（大顶堆）

    cin >> n;
    for (int i = 0; i < n; ++i) {
        scanf("%d", &arr1[i]);
    }
    for (int i = 0; i < n; ++i) {
        scanf("%d", &arr2[i]);
    }

    // 初始时，将 arr1[0] + arr2[i] 加入堆中
    for (int i = 0; i < n; ++i) {
        result.push(arr1[0] + arr2[i]);
    }

    // 然后，遍历 arr1 中的其它元素
    for (int i = 1; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            int num = arr1[i] + arr2[j];
            // 如果堆的大小超过 N，就删除堆顶元素（最大的和）
            if (result.size() < n) {
                result.push(num);
            }
            else if (num < result.top()) {
                result.pop();  // 弹出堆顶元素
                result.push(num);  // 推入新的更小的和
            }
            else
                break;
        }
    }

    // 输出最小的 N 个和
    vector<int> ans;
    while (!result.empty()) {
        ans.push_back(result.top());
        result.pop();
    }

    // 由于堆顶是最大的，输出时需要反转
    for (int i = ans.size() - 1; i >= 0; --i) {
        cout << ans[i] << " ";
    }
    cout << endl;

    return 0;
}
*/