#include <iostream>
#include <cstring>
#include <cstdio>using namespace std;
typedef long long ll;int main()
{int n;ll tmp;scanf("%d",&n);int cntodd = 0;int cnteven = 0;for(int i = 0 ; i < n; i++){scanf("%lld",&tmp);if(tmp & 1)cntodd++;elsecnteven++;}if(cntodd == 0)cout << "Second" << endl;elsecout << "First" << endl;return 0;
}

Leha like all kinds of strange things. Recently he liked the function F(n, k). Consider all possible k-element subsets of the set[1, 2, ..., n]. For subset find minimal element in it. F(n, k) — mathematical expectation of the minimal element among all k-element subsets.

But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays A and B, each consists ofm integers. For all i, j such that 1 ≤ i, j ≤ m the condition Ai ≥ Bj holds. Help Leha rearrange the numbers in the array A so that the sum  is maximally possible, where A' is already rearranged array.

First line of input data contains single integer m (1 ≤ m ≤ 2·105) — length of arrays A and B.

Next line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 109) — array A.

Next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109) — array B.

Output m integers a'1, a'2, ..., a'm — array A' which is permutation of the array A.

#include<bits/stdc++.h>using namespace std;int n;pair<int, int> a[200008];
pair<int, int> b[200008];int ansa[200008];int main() {cin >> n;for (int i = 0; i < n; i++) {cin >> a[i].first;a[i].second = i;}for (int i = 0; i < n; i++) {cin >> b[i].first;b[i].second = i;}sort(a, a + n);sort(b, b + n);for (int i = 0; i < n; i++) {ansa[b[i].second] = a[n - 1 - i].first;}for (int i = 0; i < n; i++) {printf("%d ", ansa[i]);}return 0;
}