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 n, m;
bool arr[1010];
int c[1010];
bool dp[2020];
vector<int> vec;
void zero_one(int val) {
  for (int i = 2 * m - 2; i >= val; i--) {
    dp[i] = dp[i - val] | dp[i];
  }
  dp[val] = true;
  for (int i = 2 * m - 2; i >= m; i--) {
    dp[i % m] = dp[i % m] | dp[i];
  }
}
void solve() {
  for (int i = 0; i < vec.size(); i++) {
    int k = 1;
    int p = c[vec[i]];
    while (k < p) {
      zero_one((k * vec[i]) % m);
      p -= k;
      k <<= 1;
    }
    zero_one((p * vec[i]) % m);
  }
  if (dp[0])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  scanf("%d%d", &n, &m);
  int a;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    arr[a % m] = true;
    c[a % m]++;
  }
  if (arr[0])
    cout << "YES" << endl;
  else {
    for (int i = 1; i < m; i++) {
      if (arr[i]) vec.push_back(i);
    }
    solve();
  }
  return 0;
}
