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;
long long int dp[1002][1002];
vector<long long int> v;
long long int m, res;
long long int n;
long long int ispossible(long long int index, long long int mod) {
  if (index >= n) return 0;
  if (dp[index][mod] != -1) return dp[index][mod];
  if (res) return 1;
  long long int newmod = (v.at(index) % m + mod) % m;
  if (newmod == 0) {
    res = 1;
    dp[index + 1][newmod] = 1;
    return 1;
  }
  long long int p1 = ispossible(index + 1, newmod);
  long long int p2 = ispossible(index + 1, mod);
  if (p1 || p2) {
    dp[index][mod] = 1;
    dp[index + 1][newmod] = 1;
    res = 1;
  } else {
    dp[index][mod] = 0;
    dp[index + 1][newmod] = 0;
    res = 0;
  }
  return res;
}
int main() {
  long long int i, j, a;
  scanf("%lld %lld", &n, &m);
  res = 0;
  for (i = 0; i < 1002; i++) {
    for (j = 0; j < 1002; j++) {
      dp[i][j] = -1;
    }
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &a);
    v.push_back(a % m);
  }
  if (n > m) {
    printf("YES\n");
    return 0;
  }
  ispossible(0, 0);
  if (res)
    printf("YES\n");
  else
    printf("NO\n");
}
