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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return (a % b == 0 ? b : gcd(b, a % b));
}
template <typename P>
struct Cmp {
  bool operator()(const P &p1, const P &p2) {
    if (p1.first > p2.first) return true;
    if (p1.first == p2.first) return p1.second < p2.second;
    return false;
  }
};
int main() {
  int n, m;
  scanf("%d", &n), scanf("%d", &m);
  int a[n];
  for (int i = int(0); i < int(n); i++) {
    scanf("%d", &a[i]);
  }
  vector<int> v1;
  map<int, int> m1;
  for (int i = int(0); i < int(n); i++) {
    int z = a[i] % m;
    if (z == 0) {
      puts("YES");
      return 0;
    }
    if (i == 0) {
      v1.push_back(z);
      m1[z]++;
      continue;
    }
    int sz = v1.size();
    for (int j = int(0); j < int(sz); j++) {
      int x = (z + v1[j]) % m;
      if (x == 0) {
        puts("YES");
        return 0;
      }
      if (!m1[x]) {
        v1.push_back(x);
        m1[x]++;
      }
    }
    v1.push_back(z);
  }
  puts("NO");
  return 0;
}
