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 maxn = 1000000 + 10;
const int maxm = 1000 + 10;
int presum[maxn];
bool can[maxm][maxm];
int myhash[maxm];
int a[maxn];
int n, m;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] %= m;
  }
  if (n > m) {
    presum[0] = a[0];
    myhash[presum[0]]++;
    for (int i = 1; i < n; i++) {
      presum[i] = (presum[i - 1] + a[i]) % m;
      myhash[presum[i]]++;
      if (myhash[presum[i]] == 2) {
        puts("YES");
        return 0;
      }
    }
    puts("NO");
    return 0;
  } else {
    can[0][a[0]] = true;
    can[0][0] = true;
    if (a[0] == 0) {
      puts("YES");
      return 0;
    }
    for (int i = 1; i < n; i++) {
      for (int r = 0; r < m; r++) {
        if (can[i - 1][r]) {
          if ((r + a[i]) % m == 0) {
            puts("YES");
            return 0;
          }
          can[i][r] = true;
          can[i][(r + a[i]) % m] = true;
        }
      }
    }
    puts("NO");
    return 0;
  }
  return 0;
}
