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;
int main() {
  vector<long long> v;
  long long a, b;
  cin >> a >> b;
  long long arr[a], arr2[b];
  for (int i = 0; i < a; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < b; i++) {
    cin >> arr2[i];
  }
  sort(arr, +arr + a);
  for (int i = 0; i < b; i++) {
    long long low = 0, high = a - 1, mid = (low + high) / 2, end;
    while (low <= high) {
      if (arr[mid] > arr2[i]) {
        high = mid - 1;
        mid = (low + high) / 2;
      } else if (arr[mid] <= arr2[i]) {
        end = mid;
        low = mid + 1;
        mid = (low + high) / 2;
      }
    }
    if (arr[0] <= arr2[i])
      v.push_back(end + 1);
    else
      v.push_back(0);
  }
  for (int i = 0; i < v.size(); i++) {
    if (i < v.size() - 1)
      cout << v[i] << " ";
    else
      cout << v[i];
  }
  return 0;
}
