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;
template <typename T>
vector<T> readVector(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
template <typename T>
void printVector(vector<T> in) {
  int n = in.size();
  for (int i = 0; i < n; i++) {
    cout << in[i] << " \n"[i == n - 1];
  }
}
const long long int mod = 1e9 + 7;
long long int powmod(long long int base, long long int exp,
                     long long int M = mod) {
  long long int ans = 0;
  while (exp) {
    if (exp & 1) ans = (base * ans) % M;
    base = (base * base) % M;
    exp /= 2;
  }
  return ans;
}
bool dp[1010][1010][2];
int main() {
  long long int n, m;
  cin >> n >> m;
  if (n > m) {
    return cout << "YES", 0;
  }
  auto t = readVector<long long int>(n);
  dp[0][(t[0]) % m][1] = 1;
  dp[0][0][0] = 1;
  for (long long int i = 1; i < n; i++) {
    for (long long int j = 0; j <= m; j++) {
      if (dp[i - 1][j][0]) {
        dp[i][j][0] = 1;
        dp[i][(j + t[i] % m) % m][1] = 1;
      }
      if (dp[i - 1][j][1]) {
        dp[i][j][1] = 1;
        dp[i][(j + t[i] % m) % m][1] = 1;
      }
    }
  }
  if (dp[n - 1][0][1])
    puts("YES");
  else
    puts("NO");
}
