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;
long long int ara[1000005], n, m, dp[1005][1005];
long long int func(long long int i, long long int j) {
  if ((j + ara[i]) % m == 0) {
    return 1;
  }
  if (i >= n - 1) {
    if ((j + ara[i]) % m == 0) {
      return 1;
    } else
      return 0;
  }
  if (dp[i][j] != -1) {
    return dp[i][j];
  }
  long long int p1 = 0, p2 = 0;
  p1 = func(i + 1, (j + ara[i]) % m);
  p2 = func(i + 1, j);
  dp[i][j] = p1 | p2;
  return dp[i][j];
}
int main() {
  long long int i, k, sum = 0;
  scanf("%I64d %I64d", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%I64d", &ara[i]);
  }
  if (n > m) {
    printf("YES\n");
    return 0;
  }
  memset(dp, -1, sizeof dp);
  k = func(0, 0);
  if (k == 1) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}
