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;
const int N = 1e6 + 10;
const int M = 1e3 + 10;
bool f[M], g[M];
int b[M];
int c[20];
int d[20];
int main() {
  c[0] = 1;
  for (int i = 1; i < 20; i++) c[i] = c[i - 1] << 1;
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(b, 0, sizeof(b));
    for (int i = 0, x; i < n; i++) {
      scanf("%d", &x);
      b[x % m]++;
      if (x % m == 0) {
        printf("YES\n");
        return 0;
      }
    }
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    f[0] = 1;
    bool flag = 0;
    for (int i = 1; i < m; i++) {
      int top = 0;
      while (b[i]) {
        if (b[i] >= c[top])
          d[top] = c[top];
        else
          d[top] = b[i];
        b[i] -= d[top];
        top++;
      }
      for (int j = 0; j < top; j++) {
        int y = d[j] * i;
        for (int k = m; k >= 0; k--) {
          if (f[k]) {
            g[(k + y) % m] = 1;
          }
        }
        if (g[0]) {
          flag = 1;
          break;
        }
        for (int k = 0; k <= m; k++) {
          if (g[k]) f[k] = 1;
        }
      }
    }
    printf("%s\n", flag ? "YES" : "NO");
  }
  return 0;
}
