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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
using namespace std;
int dp[1000 + 5];
int a[1000004];
int dp1[1000 + 5];
int main() {
  int n, m;
  int i, j;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] %= m;
  }
  dp[0] = 1;
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      dp1[j] = 0;
      int z = (j - a[i]) % m;
      z = (z + m) % m;
      if (dp[z] != 0) dp1[j] = 1;
    }
    if (dp1[m]) {
      cout << "YES";
      return 0;
    }
    for (j = 0; j <= m; j++) dp[j] = (dp[j] || dp1[j]);
  }
  if (!dp[m])
    printf("NO");
  else
    printf("YES");
  return 0;
}
