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 long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  long long int g = (a * b) / gcd(a, b);
  return g;
}
const int N = 998244353;
const int mod = 1e8;
void swap(char *p, char *q) {
  char temp = *p;
  *p = *q;
  *q = temp;
  return;
}
long long int powxy(long long int x, long long int y) {
  if (y == 0) return 1;
  if (y % 2 == 1) return (x * powxy(x, y - 1)) % N;
  long long int t = powxy(x, y / 2);
  return (t * t) % N;
}
int dp[1001][1001];
int a[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= k;
  }
  if (n >= k) {
    cout << "YES";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    dp[i][a[i - 1]] = 1;
    for (int j = 0; j < k; j++) {
      dp[i][j] |= dp[i - 1][j];
      dp[i][(j + a[i - 1]) % k] |= dp[i - 1][j];
    }
  }
  if (dp[n][0])
    cout << "YES";
  else
    cout << "NO";
}
