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 inf = 1e9 + 10;
const long long INF = 4e18 + 10;
const int maxn = 1e6 + 5;
const int maxm = 1e3 + 5;
int n, m;
int cnt[maxm];
set<int> bag[maxm];
bool d[maxm][maxm];
int MAIN() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    int ai;
    scanf("%d", &ai);
    if (ai % m == 0) {
      puts("YES");
      return 0;
    }
    ++cnt[ai % m];
  }
  if (cnt[0] > 0) {
    puts("YES");
    return 0;
  }
  for (int x = 0; x < m; ++x) {
    int s = 0;
    for (int i = 0; i < cnt[x]; ++i) {
      s += x;
      bag[x].insert(s % m);
      d[x][s % m] = true;
    }
  }
  for (int x = 1; x + 1 < m; ++x) {
    for (int r = 0; r < m; ++r) {
      if (d[x][r]) {
        d[x + 1][r] = true;
        for (set<int>::iterator it = bag[x + 1].begin(); it != bag[x + 1].end();
             ++it) {
          d[x + 1][(r + *it) % m] = true;
        }
      }
    }
  }
  for (int x = 1; x < m; ++x) {
    if (d[x][0]) {
      puts("YES");
      return 0;
    }
  }
  puts("NO");
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  int ret = MAIN();
  return ret;
}
