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;
int n, m, a[1000006], i, j;
bool dp[1004][2], x;
int main() {
  memset(dp, 0, sizeof(dp));
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] %= m;
    if (a[i] == 0) {
      printf("YES\n");
      return 0;
    }
  }
  for (i = 1; i < m; i++) dp[i][0] = 0;
  dp[0][0] = 1;
  for (j = 1; j <= n; j++) {
    for (i = 0; i < m; i++) {
      x = dp[(i - a[j - 1] + m) % m][(j - 1) & 1];
      dp[i][j & 1] = dp[i][(j - 1) & 1] || x;
      if (i == 0 && x) {
        printf("YES\n");
        return 0;
      }
    }
  }
  printf("NO\n");
  return 0;
}
