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 dp[1000 + 10][1000 + 10][2];
int arr[1000 + 10];
int n, m, k, d;
int rec(int index, int mo, int flag) {
  if (index == 1001 && mo != 0) return 0;
  if (mo == 0 && flag) {
    printf("YES");
    exit(0);
  }
  int &st = dp[index][mo][flag];
  if (st != -1) return 0;
  if (arr[index] == 0) {
    rec(index + 1, mo, flag);
  } else {
    for (int i = 0; i <= arr[index]; i++) {
      if (i > 0) flag = 1;
      rec(index + 1, (mo % m + (i * index) % m) % m, flag);
    }
  }
  dp[index][mo][flag] = 0;
}
int main() {
  memset(dp, -1, sizeof dp);
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &k);
    arr[k % m]++;
    if (k == 0) {
      printf("YES\n");
      return 0;
    }
    if (k % m == 0) {
      printf("YES");
      return 0;
    }
  }
  rec(1, 0, 0);
  printf("NO");
  return 0;
}
