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, f[2][1010], a[1000010], k;
int sub(int x, int y) {
  int t = (x - y) % k;
  if (t > 0)
    return t;
  else
    return t + k;
}
void process2() {
  for (int t = 0; t < k; ++t) f[0][t] = 2e9;
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i)
    for (int t = 0; t < k; ++t) {
      f[i % 2][t] = 2e9;
      if (f[1 - i % 2][t] < f[1 - i % 2][sub(t, a[i])] + 1)
        f[i % 2][t] = f[1 - i % 2][t];
      else
        f[i % 2][t] = f[1 - i % 2][sub(t, a[i])] + 1;
    }
  int sum = 0;
  for (int i = 1; i <= n; ++i) sum = (sum + a[i]) % k;
  if (f[n % 2][sum] >= n)
    printf("NO");
  else
    printf("YES");
}
int s[1000010] = {0};
set<int> S;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  if (n <= k)
    process2();
  else {
    for (int i = 1; i <= n; ++i) s[i] = (s[i - 1] + a[i]) % k;
    S.insert(s[0]);
    for (int i = 1; i <= n; ++i) {
      if (S.find(s[i]) != S.end()) {
        printf("YES");
        return 0;
      }
      S.insert(s[i]);
    }
    printf("NO");
  }
  return 0;
}
