You are given a sequence of numbers a1, a2, ..., an, and a number m.

Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

Input

3 5
1 2 3


Output

YES


Input

1 6
5


Output

NO


Input

4 6
3 1 1 3


Output

YES


Input

6 6
5 5 5 5 5 5


Output

YES

Note

In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.
#include <bits/stdc++.h>
using namespace std;
bool mark[2][1111];
int main() {
  int a, b, c, d, e, f, g, h;
  scanf("%d%d", &a, &b);
  if (a >= b) {
    printf("YES\n");
    return 0;
  }
  mark[0][0] = 1;
  for (f = d = g = 0; d < a; d++, f ^= 1) {
    scanf("%d", &e);
    e %= b;
    if (e == 0) {
      printf("YES\n");
      return 0;
    }
    for (c = b - 1; c > -1; c--) {
      if (mark[f][c] && c + e == b) {
        printf("YES\n");
        return 0;
      }
      if (mark[f][c]) {
        mark[f ^ 1][(c + e) % b] = 1;
      }
    }
    for (c = 0; c < b; c++) {
      if (mark[f][c]) mark[f ^ 1][c] = 1;
    }
  }
  printf("NO\n");
  return 0;
}
