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;
string vow = "aeiou";
int month[] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const long double pie = 3.14159265358979;
const long long mod = 1e9 + 7;
int n, m;
vector<int> v;
void good() {
  cout << "YES";
  cout << "\n";
  ;
  exit(0);
}
void read() {
  cin >> n >> m;
  if (n >= m) good();
  v.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i] %= m;
  }
}
const int N = 1e3 + 2, inf = 1e5;
bool vis[N][N];
void f(int indx, int rem) {
  if (indx < 0) return;
  if (vis[indx][rem]) return;
  vis[indx][rem] = 1;
  f(indx - 1, rem);
  rem += v[indx];
  rem %= m;
  if (rem == 0) good();
  f(indx - 1, rem);
}
void solve(int test_case) {
  read();
  f(n - 1, 0);
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) solve(i);
  return 0;
}
