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 = (int)1e9;
const int mod = (int)1e9 + 7;
const int MAX = (int)1e6 + 5;
long long store[MAX];
int dp[1005][1005];
int main(void) {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = (1); i < (n + 1); ++i) scanf("%lld", store + i);
  if (n > m)
    puts("YES");
  else {
    dp[1][store[1] % m] = 1;
    for (int i = (2); i < (n + 1); ++i) {
      dp[i][store[i] % m] = 1;
      for (int j = 0; j < (m); ++j)
        if (dp[i - 1][j]) {
          dp[i][j] = 1;
          dp[i][(j + store[i] % m) % m] = 1;
        }
    }
    for (int i = (1); i < (n + 1); ++i)
      if (dp[i][0]) {
        puts("YES");
        return 0;
      }
    puts("NO");
  }
  return 0;
}
