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 M = 1002;
int cnt[M];
int dp[2][M];
int main() {
  int i, j, n, m, ai;
  int two, am, flag = 0;
  int z = 0;
  scanf("%d%d", &n, &m);
  fill(cnt, cnt + m, 0);
  fill(dp[0], dp[0] + m, 0);
  dp[0][0] = 1;
  fill(dp[1], dp[1] + m, 0);
  dp[1][0] = 1;
  for (i = 0; i < n; i++) {
    scanf("%d", &ai);
    cnt[ai % m]++;
    if (ai == 0) z++;
  }
  if (z) {
    puts("Yes");
    return 0;
  }
  for (i = 0; i < m; i++) {
    if (!cnt[i]) continue;
    two = 1;
    while (cnt[i] >= two) {
      am = (two * i) % m;
      for (j = 0; j < m; j++) dp[flag][j] = dp[flag ^ 1][j];
      for (j = 0; j < m; j++) dp[flag][(j + am) % m] += (dp[flag ^ 1][j] > 0);
      cnt[i] -= two;
      two <<= 1;
      flag ^= 1;
    }
    if (cnt[i]) {
      am = (cnt[i] * i) % m;
      for (j = 0; j < m; j++) dp[flag][j] = dp[flag ^ 1][j];
      for (j = 0; j < m; j++) dp[flag][(j + am) % m] += (dp[flag ^ 1][j] > 0);
      flag ^= 1;
    }
  }
  puts(dp[flag ^ 1][0] > 1 ? "YES" : "NO");
  return 0;
}
