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 long long N = 1000010;
vector<int> v;
long long sum[N];
map<int, int> val;
bool falt = false;
int dp[1001][1001];
int n, m;
void calc(int i, int j) {
  if (j % m == 0 && j != 0) falt = true;
  if (i >= n) return;
  if (dp[i][j] != -1) return;
  if ((j + v[i]) % m == 0) falt = true;
  calc(i + 1, (j + v[i]) % m);
  calc(i + 1, j);
  dp[i][j] = 1;
}
int main() {
  int x;
  long long total = 0, d = 0;
  bool ans = false;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x);
    v.push_back(x);
    if (x % m == 0) ans = true;
    if (val[x] == 0) d++;
    val[x]++;
  }
  if (d >= m || n >= 1000) ans = true;
  if (!ans) {
    memset(dp, -1, sizeof(dp));
    calc(0, 0);
  }
  if (falt) ans = true;
  if (ans)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
