You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.

Input

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

The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).

The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).

Output

Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.

Examples

Input

5 4
1 3 5 7 9
6 4 2 8


Output

3 2 1 4


Input

5 5
1 2 1 2 5
3 1 4 1 5


Output

4 2 4 2 5
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 19;
int n, m;
int C[maxn << 1];
int sample[maxn << 1];
int a[maxn], b[maxn];
void add(int x, int d) {
  while (x <= n + m) {
    C[x] += d;
    x += x & -x;
  }
}
int sum(int x) {
  int ret = 0;
  while (x) {
    ret += C[x];
    x -= x & -x;
  }
  return ret;
}
int main() {
  while (cin >> n >> m) {
    memset(C, 0, sizeof(C));
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      sample[i] = a[i];
    }
    for (int i = 0; i < m; i++) {
      scanf("%d", &b[i]);
      sample[n + i] = b[i];
    }
    sort(sample, sample + m + n);
    int len = unique(sample, sample + m + n) - sample;
    for (int i = 0; i < n; i++) {
      a[i] = lower_bound(sample, sample + len, a[i]) - sample + 1;
      add(a[i], 1);
    }
    for (int i = 0; i < m; i++) {
      b[i] = lower_bound(sample, sample + len, b[i]) - sample + 1;
    }
    for (int i = 0; i < m; i++) {
      printf("%d%c", sum(b[i]), " \n"[i == m - 1]);
    }
  }
  return 0;
}
