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;
map<long long int, long long int> g;
long long int pre[1000005] = {0}, a[1000005];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = a[i] % m;
    if (a[i] == 0) {
      cout << "YES\n";
      return 0;
    }
  }
  if (n > m) {
    pre[0] = 0;
    g[0] = 1;
    for (long long int i = 0; i < n; i++) {
      pre[i + 1] = (pre[i] + a[i]) % m;
      g[pre[i + 1]]++;
    }
    map<long long int, long long int>::iterator it;
    for (it = g.begin(); it != g.end(); ++it) {
      long long int x = it->second;
      if (x >= 2) {
        cout << "YES\n";
        return 0;
      }
    }
    cout << "NO\n";
    return 0;
  } else {
    long long int dp[2][m + 2];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    long long int x = 0;
    long long int cnt = 0;
    while (cnt < n) {
      x ^= 1;
      dp[x][0] = 1;
      dp[x][a[cnt]] = 1;
      for (long long int j = 1; j <= m; j++) {
        dp[x][j] |= dp[x ^ 1][j];
        if (dp[x ^ 1][j] == 1) {
          dp[x][(j + a[cnt]) % m] = 1;
          if ((j + a[cnt]) % m == 0) {
            cout << "YES\n";
            return 0;
          }
        }
      }
      cnt++;
    }
    cout << "NO\n";
    return 0;
  }
}
