Codeforces Round #429 (Div. 2)：C、&840A、 Leha and Function

题目：

Leha like all kinds of strange things. Recently he liked the functionF(n, k). Consider all possiblek-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 allk-element subsets.

But only function does not interest him. He wants to do interesting things with it. Mom brought him two arraysA and B, each consists ofm integers. For all i, j such that 1 ≤ i, j ≤ m the conditionA_i ≥ B_j holds. Help Leha rearrange the numbers in the arrayA 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 arraysA and B.

Next line contains m integersa₁, a₂, ..., a_m (1 ≤ a_i ≤ 10⁹) — array A.

Next line contains m integersb₁, b₂, ..., b_m (1 ≤ b_i ≤ 10⁹) — array B.

Output

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

Examples

Input

57 3 5 3 42 1 3 2 3

Output

4 7 3 5 3

Input

74 6 5 8 8 2 62 1 2 2 1 1 2

Output

2 6 4 5 8 8 6

题意：给出a、b两个数组，a数组任意一个数都大于b数组的任意一个数，将a数组重新排列使得 最大

思路：题目有给出提示：定义F(n,k)：F(n, k)是在集合{1, 2, 3, ..., n}中所有的具有k个元素的子集中分别取最小值，相加后的期望。For example, let's find F(4, 2). All possible 2-element subsets of {1,2,3,4} are: {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. Their minimal values are 1, 1, 1, 2, 2, 3. So the average (expected) value is F(4,2)=(1+1+1+2+2+3)/6=10/6=1.66666666

可得两个结论：

F(n,k)>F(nm,k)   n>m

F(n,k)>F(n,p)   k<p

所以a中最小的数对应b中最大的数，依次递推

code：

#include<stdio.h>#include<algorithm>using namespace std;int a[200005],b[200005];struct node{        int x,pos;}q[200005];int cmp(node t1,node t2){        return t1.x>t2.x;}int main(){        int n,i;        while(~scanf("%d",&n)){                for(i=0;i<n;i++) scanf("%d",&a[i]);                for(i=0;i<n;i++){                        scanf("%d",&q[i].x);                        q[i].pos=i;                }                sort(a,a+n);                sort(q,q+n,cmp);                for(i=0;i<n;i++){                        b[q[i].pos]=a[i];                }                for(i=0;i<n;i++) printf("%d%c",b[i],i==n?'\n':' ');        }        return 0;}