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 N = 1e3 + 100;
int n, Mod;
int Dp[N][N][2];
vector<pair<int, int>> Arr;
int Solve(int idx, int Sum, int taken) {
  if (idx == n) {
    if (!Sum && taken) {
      cout << "YES\n";
      exit(0);
    }
    return 0;
  }
  int &ret = Dp[idx][Sum][taken];
  if (ret + 1) return ret;
  ret = Solve(idx + 1, Sum, taken);
  for (int i = 1; i <= Arr[idx].second; i++)
    ret = max(ret, Solve(idx + 1, (Sum + i * Arr[idx].first) % Mod, 1));
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> Mod;
  map<int, int> Cnt;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    Cnt[x % Mod]++;
  }
  for (auto x : Cnt) {
    vector<bool> vis(Mod, false);
    int cur = 0;
    for (int i = 0; i < x.second; i++) {
      cur = (cur + x.first) % Mod;
      if (vis[cur]) {
        Cnt[x.first] = i;
        if (cur == 0) return cout << "YES\n", 0;
        break;
      } else
        vis[cur] = true;
    }
  }
  for (auto x : Cnt) Arr.push_back({x.first, x.second});
  n = Arr.size();
  memset(Dp, -1, sizeof Dp);
  if (Solve(0, 0, 0))
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
