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 mod = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  if (a >= mod) a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    if (res >= mod) res %= mod;
    a = a * a;
    if (a >= mod) a %= mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int ok[1005] = {0};
int main() {
  int i, j, t;
  int z, n, m, x, y;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    x = x % m;
    for (j = 0; j < m; j++) {
      if (ok[j] == 0 || ok[j] == i) continue;
      z = x + j;
      if (z >= m) z -= m;
      if (ok[z] == 0) ok[z] = i;
    }
    ok[x] = i;
    if (ok[0]) {
      puts("YES\n");
      return 0;
    }
  }
  puts("NO\n");
  return 0;
}
