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>
#pragma GCC optimize("O3")
using namespace std;
long long max3(long long x, long long y, long long z) {
  return max(max(x, y), z);
}
long long min3(long long x, long long y, long long z) {
  return min(min(x, y), z);
}
const long long M = 2e5 + 10, M2 = 1e6 + 10, mod = 1e9 + 7, inf = 1e17 + 10;
void add(int& a, int b) {
  a += b;
  if (a >= mod) {
    a -= mod;
  }
}
int X[] = {0, 1, 0, -1};
int Y[] = {-1, 0, 1, 0};
long long power(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = ((x % mod) * (x % mod)) % mod;
    n = n / 2;
  }
  return result;
}
long long n, m;
long long a[1000001];
long long dp[1001][1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i < n + 1; i++) {
    cin >> a[i];
    a[i] = a[i] % m;
  }
  for (long long i = 1; i < n + 1; i++) {
    if (dp[i - 1][0] == 1) {
      cout << "YES" << endl;
      return 0;
    }
    dp[i][a[i]] = 1;
    for (long long j = 0; j < m; j++) {
      if (dp[i - 1][j] == 1) {
        dp[i][(j + a[i]) % m] = 1;
        dp[i][j] = 1;
      }
    }
  }
  if (dp[n][0] == 1) {
    cout << "YES" << endl;
    return 0;
  }
  cout << "NO" << endl;
  return 0;
}
