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 = 1e5 + 5;
const int M = 1e6 + 5;
const int MOD = 1e9 + 7;
const int CM = 998244353;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
template <typename T>
T MAX(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
T MIN(T a, T b) {
  return a > b ? b : a;
}
template <typename T>
T GCD(T a, T b) {
  return b == 0 ? a : GCD(b, a % b);
}
template <typename T>
T LCM(T a, T b) {
  return a / GCD(a, b) * b;
}
template <typename T>
T ABS(T a, T b) {
  return a > 0 ? a : -a;
}
template <typename T>
T ADD(T a, T b, T MOD) {
  return (a + b) % MOD;
}
template <typename T>
T DEL(T a, T b, T MOD) {
  return ((a - b) % MOD + MOD) % MOD;
}
std::set<int> vx;
void solve() {
  int n, m;
  cin >> n >> m;
  if (n >= m) {
    printf("YES\n");
    return;
  }
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    std::vector<int> v;
    for (auto x : vx) {
      v.push_back(x);
    }
    vx.insert(a % m);
    for (auto x : v) {
      vx.insert((x + a) % m);
    }
  }
  if (vx.count(0) != 0)
    printf("YES\n");
  else
    printf("NO\n");
}
void TIME() {
  clock_t start, finish;
  double totaltime;
  start = clock();
  solve();
  finish = clock();
  totaltime = (double)(finish - start) / CLOCKS_PER_SEC;
  printf("\nTime:%lfms\n", totaltime * 1000);
}
int main() {
  solve();
  return 0;
}
