codeforces 841C Leha and Function

C. Leha and Function

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 mintegers. 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.

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

題意：題目一大堆英文，看不懂，但是根據樣例可以猜出題意，最小的Bi對應最大的Ai，最大的Bi對應最小的Ai，依此類推

建一個結構體數組，一個保存原數組序列，第二個保存當前數的數值，排序模擬一下就好了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define epp tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
const int Max=2e5+5;
int a[Max+1],b[Max+1],m,c[Max+1];

struct node
{
    int x,y;
}e[Max+1];

bool cmp(const node& a,const node& b)
{
    return a.y<b.y;
}

bool cmp1(int x,int y)
{
    return x>y;
}

int main()
{
    while(~sf("%d",&m))
    {
        for0(i,m)
            sf("%d",&a[i]);
        for0(i,m)
            sf("%d",&e[i].y),e[i].x=i;
        sort(a,a+m,cmp1);
        sort(e,e+m,cmp);
        for0(i,m)
            b[e[i].x]=i;//x表示原序列下標，i表示排序後的下標
        for0(i,m)
            pf(i==m-1?"%d\n":"%d ",a[b[i]]);
    }
    return 0;
}