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;
int dr[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dc[] = {1, -1, 0, 0, -1, 1, -1, 1};
int const MAX = 1e6 + 7;
int const MOD = 1e9 + 7;
int n, m, arr[MAX], mem[1007][1007][2];
bool dp(int i, int m_sum, int x) {
  if (!m_sum && x) return 1;
  if (i == n) return 0;
  if (mem[i][m_sum][x] != -1) return mem[i][m_sum][x];
  return mem[i][m_sum][x] =
             dp(i + 1, (m_sum + arr[i]) % m, 1) | dp(i + 1, m_sum, x);
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  if (n > m) return (cout << "YES\n", 0);
  memset(mem, -1, sizeof mem);
  if (dp(0, 0, 0)) return (cout << "YES\n", 0);
  return (cout << "NO\n", 0);
}
