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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int hashs[1000005], a[1000005 * 3];
bool dp[3][1000005];
int main() {
  memset(hashs, 0, sizeof(hashs));
  int flag = 0;
  int n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = a[i] % m;
    if (a[i] == 0) a[i] = m;
  }
  dp[0][0] = true;
  dp[1][0] = true;
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 0; j--) {
      if (dp[1][j]) {
        if (j + a[i] == m) {
          puts("YES");
          return 0;
        }
        dp[0][(j + a[i]) % m] = true;
      }
    }
    memcpy(dp[1], dp[0], sizeof(dp[0]));
  }
  cout << "NO" << endl;
  return 0;
}
