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 INF = 0x3f3f3f3f;
const int N = 1000010;
int i, j, k;
int m;
int res, ans, cnt, temp;
int a[1000005];
int sum[1005];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  if (n > m) {
    printf("YES\n");
    return 0;
  }
  set<int> s;
  set<int>::iterator it;
  s.insert(a[0] % m);
  for (i = 1; i < n; i++) {
    int l = -1;
    it = s.begin();
    int si = *it;
    if (si == 0) {
      printf("YES\n");
      return 0;
    }
    for (it = s.begin(); it != s.end(); ++it) {
      l++;
      int si = *it;
      sum[l] = (si + a[i]) % m;
    }
    for (j = 0; j <= l; j++) s.insert(sum[j]);
    s.insert(a[i]);
  }
  it = s.begin();
  int si = *it;
  if (si == 0) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}
