codeforces Queries about less or equal elements 二分

   You are given two arrays of integers a andb. For each element of the second arrayb_j you should find the number of elements in arraya that are less than or equal to the valueb_j.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·10⁵) — the sizes of arraysa and b.

The second line contains n integers — the elements of arraya ( - 10⁹ ≤ a_i ≤ 10⁹).

The third line contains m integers — the elements of arrayb ( - 10⁹ ≤ b_j ≤ 10⁹).

Output

Print m integers, separated by spaces: thej-th of which is equal to the number of such elements in arraya that are less than or equal to the valueb_j.

Sample test(s)

Input

5 41 3 5 7 96 4 2 8

Output

3 2 1 4

Input

5 51 2 1 2 53 1 4 1 5

Output

4 2 4 2 5

code

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<stack>#include<queue>#include<fstream>#include <functional>#include <iomanip>#include<set>#include<vector>#include<map>#include <complex>#include <bitset>#define lson l,m,rt<<1#define lson m+1,r,rt<<1|1#define Max(a,b) a>b?a:b#define Min(a,b) a<b?a:b#define esp 1e-6#define LL long longusing namespace std;int a[200007],b[200007];int query(int l,int r,int j){    int p=l,q=r;    while(p<=q)    {        int m=q+((p-q)>>1);        if(b[j]>=a[m])        p=m+1;        else        q=m-1;    }    return p-1;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        sort(a,a+n);        int c[200007]={0};        for(int j=0; j<m; j++)        {            scanf("%d",&b[j]);           c[j]=query(0,n-1,j)+1;        }        for(int i=0;i<m;i++)            printf("%d ",c[i]);        printf("\n");    }}