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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 1005;
long long dp[N][N];
bool vis[N][N];
long long a[N];
long long n, m;
bool calc(long long level, long long num) {
  if (num == 0) return true;
  if (level == m) {
    return false;
  }
  if (vis[level][num]) return false;
  vis[level][num] = true;
  if (calc(level + 1, num)) return true;
  for (long long i = 1; i <= a[level]; i++) {
    long long temp = (num - i * level) % m;
    temp = (temp + m) % m;
    if (calc(level + 1, temp)) {
      return true;
    }
  }
  return false;
}
int main() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    x = x % m;
    a[x]++;
  }
  if (a[0] > 0) {
    cout << "YES" << endl;
    return 0;
  }
  if (calc(1, m))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
