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;
int gcd(int a, int b) {
  int r, i;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long int a[1000100];
long long int b[1000100];
long long int c[10001];
int main() {
  long long int i, j, n, m, k, x;
  scanf("%lld", &n);
  scanf("%lld", &m);
  set<long long int> s1;
  map<long long int, long long int> mp;
  int f = 0;
  for (i = (0); i < (n); i++) {
    scanf("%lld", &a[i]);
    k = a[i] % m;
    if (k == 0) {
      f = 1;
    }
    if (f == 0) {
      if (i == 0) {
        s1.insert(k);
        mp[k] = 1;
      } else {
        long long int ar[1001];
        j = 0;
        long long int v1 = a[i] % m;
        if (v1 == 0) {
          f = 1;
        } else {
          ar[j++] = v1;
        }
        set<long long int>::iterator it;
        for (it = s1.begin(); it != s1.end(); it++) {
          long long int po = *it;
          long long int vv = (po + a[i]) % m;
          if (vv == 0) {
            f = 1;
            break;
          }
          ar[j++] = vv;
        }
        if (f == 0) {
          int i1;
          for (i1 = 0; i1 < j; i1++) {
            s1.insert(ar[i1]);
            mp[ar[i1]] = 1;
          }
        }
      }
    }
  }
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
