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 n, m;
int arr[1230122];
bool dp[2000][2000];
bool vis[2000][2000];
bool solve(int index, int mod) {
  if (vis[index][mod]) return dp[index][mod];
  vis[index][mod] = true;
  if (index == n) return dp[index][mod] = false;
  int outro = (mod + arr[index]) % m;
  if (outro == 0) return dp[index][mod] = true;
  return dp[index][mod] = solve(index + 1, outro) || solve(index + 1, mod);
}
int main() {
  scanf("%d %d", &n, &m);
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    arr[i] %= m;
  }
  if (n > m) {
    printf("YES\n");
  } else {
    printf("%s\n", solve(0, 0) ? "YES" : "NO");
  }
  return 0;
}
