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;
const int M = 1e9 + 7;
int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
int expo(int x, int y) {
  int res = 1;
  x = x % M;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % M;
    y = y >> 1;
    x = (1ll * x * x) % M;
  }
  return res;
}
const int N = 1e3 + 5;
int n, m, a[1000005];
int dp[N][N];
int rec(int s, int mod) {
  if (mod == 0) return 1;
  if (s >= n) return 0;
  if (dp[s][mod] != -1) return dp[s][mod];
  dp[s][mod] = 0;
  dp[s][mod] |= rec(s + 1, (mod - a[s] + m) % m);
  dp[s][mod] |= rec(s + 1, mod);
  return dp[s][mod];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      a[i] %= m;
    }
    if (n > m) return cout << "YES", 0;
    memset(dp, -1, sizeof dp);
    int chk = rec(0, m);
    if (chk)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}
