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>
int N;
int M;
inline int MOD(const int x) {
  if (x >= M) return x - M;
  return x;
}
char d[1000 * 1000][1000 / 8 + 1];
int v[1000 * 1000];
int k, div8, mod8;
inline char get_i_j(const int i, const int j) {
  k = j;
  div8 = (k >> 3);
  mod8 = (k & 0x07);
  return d[i][div8] & (1 << mod8);
}
inline void set_i_j(const int i, const int j) {
  k = j;
  div8 = (k >> 3);
  mod8 = (k & 0x07);
  d[i][div8] |= (1 << mod8);
}
int main() {
  scanf("%d %d", &N, &M);
  if (N > M) {
    printf("YES\n");
    return 0;
  }
  for (int i = 0; i < N; ++i) {
    long t;
    scanf("%ld", &t);
    v[i] = t % M;
  }
  memset(d, 0, sizeof(d) * sizeof(char));
  set_i_j(0, v[0]);
  for (int i = 1; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      set_i_j(i, v[i]);
      if (!get_i_j(i - 1, j)) {
        continue;
      }
      set_i_j(i, j);
      set_i_j(i, MOD(j + v[i]));
    }
  }
  if (get_i_j(N - 1, 0)) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}
