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 of m 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 <image> is maximally possible, where A' is already rearranged array.

Input

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

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

Examples

Input

5
7 3 5 3 4
2 1 3 2 3


Output

4 7 3 5 3


Input

7
4 6 5 8 8 2 6
2 1 2 2 1 1 2


Output

2 6 4 5 8 8 6
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2 * 1000 * 100 + 1000;
int n;
pair<int, int> a[MAXN], b[MAXN];
int ans[MAXN];
int p[MAXN];
bool comp(const pair<int, int>& x, const pair<int, int>& y) {
  return (x.first >= y.first);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    p[i] = x;
    a[i] = make_pair(x, i);
  }
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    b[i] = make_pair(x, i);
  }
  sort(a + 1, a + n + 1);
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) {
    ans[b[i].second] = a[n - i + 1].second;
  }
  for (int i = 1; i <= n; i++) cout << p[ans[i]] << " ";
  return 0;
}
