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 double pi = 2.0 * acos(0.0);
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
long long int Dp[1005][1005] = {0};
long long int A[1000005];
long long int x, n;
int fun(int ind, int sum) {
  if (sum == x) {
    return 1;
  }
  sum %= x;
  if (ind == n) {
    return 0;
  }
  if (Dp[ind][sum] != -1) {
    return Dp[ind][sum];
  }
  int a, b;
  a = fun(ind + 1, sum + A[ind]);
  b = fun(ind + 1, sum);
  return Dp[ind][sum] = a + b;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, c, i, j, k, first, r, y, z;
  long long int m, p, q, t;
  memset(Dp, -1, sizeof(Dp));
  first = 0;
  r = 0;
  cin >> n >> x;
  for (i = 0; i < n; i++) {
    cin >> A[i];
    A[i] %= x;
  }
  for (i = 0; i < n; i++)
    if (A[i] == 0) first = 1;
  if (n > x + 3 || first == 1) {
    cout << "YES\n";
  } else {
    r = fun(0, 0);
    if (r == 0) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
    }
  }
  return 0;
}
