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(int argc, char* argv[]) {
  std::ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  vector<int> b(m);
  for (int i = 0; i < a.size(); ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < b.size(); ++i) {
    cin >> b[i];
  }
  map<int, int> d;
  for (int i = 0; i < a.size(); ++i) {
    d[a[i]]++;
  }
  int c = 0;
  for (auto i = d.begin(); i != d.end(); ++i) {
    int curr = i->second;
    ;
    c += curr;
    i->second = c;
  }
  for (int i = 0; i < b.size(); ++i) {
    auto f = d.lower_bound(b[i]);
    if (f != d.end()) {
      if (f->first != b[i]) {
        if (f != d.begin()) {
          f--;
        } else {
          cout << "0 ";
          continue;
        }
      }
      cout << f->second << ' ';
    } else {
      cout << c << ' ';
    }
  }
  cout << endl;
  set<int> s;
  return 0;
}
