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 maxn = 1e3;
int num[maxn + 10];
bool vis[maxn + 10][maxn + 10][2];
int n, m;
vector<int> a;
struct node {
  int val, index;
  bool take;
  node(int a, int b, bool c) : val(a), index(b), take(c) {}
};
string solve() {
  queue<node> q;
  q.push(node(0, 0, 0));
  while (!q.empty()) {
    int val = q.front().val;
    int index = q.front().index;
    bool take = q.front().take;
    q.pop();
    if (val == 0 && take) return "YES";
    if (index >= a.size()) continue;
    for (int i = 0; i <= num[a[index]]; i++) {
      int nx = (val + 1LL * i * a[index]) % m;
      if (!vis[nx][index][take || i]) {
        vis[nx][index][take || i] = 1;
        q.push(node(nx, index + 1, (take || i)));
      }
    }
  }
  return "NO";
}
int main() {
  scanf("%d%d", &n, &m);
  a.resize(n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    num[x % m]++;
    a[i] = x % m;
  }
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  cout << solve() << endl;
  return 0;
}
