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 = 1e6 + 11, mod = 1e9 + 7;
const int INF1 = 2e9 + 11;
const long long INF2 = 2e18 + 11;
const int base = 500;
const int P = 31;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
const double EPS = 1e-12;
const double PI = acos(-1.0);
int n, m, a[N];
bool us[N];
vector<int> v;
int main() {
  if (fopen("queue"
            ".in",
            "r"))
    freopen(
        "queue"
        ".in",
        "r", stdin),
        freopen(
            "queue"
            ".out",
            "w", stdout);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i += 1) {
    cin >> a[i];
    int t = v.size();
    for (auto& it : v)
      if (!us[(it + a[i]) % m])
        us[(it + a[i]) % m] = 1, v.emplace_back((it + a[i]) % m);
    if (!us[a[i] % m]) v.emplace_back(a[i] % m), us[a[i] % m] = 1;
    if (us[0]) return !(cout << "YES");
  }
  cout << "NO";
  return 0;
}
