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>
const int MAX_L = 2005;
const int INF_INT = 0x3f3f3f3f;
const long long INF_LL = 0x7fffffff;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const double pi = acos(-1);
using namespace std;
int n, m;
int a[1000100];
int vis[1010];
int main() {
  while (cin >> n >> m) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      scanf("%d", a + i);
      a[i] %= m;
    }
    int flag = 0;
    vis[0] = 1;
    int temp[1010];
    for (int i = 1; i <= n; i++) {
      memset(temp, 0, sizeof(temp));
      for (int j = 0; j < m; j++)
        if (vis[j]) {
          int tmp = a[i] + j;
          if ((tmp % m) == 0) {
            flag = 1;
            break;
          }
          temp[tmp % m] = 1;
        }
      for (int j = 0; j < m; j++)
        if (temp[j]) vis[j] = 1;
      if (flag) break;
    }
    if (flag)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
