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>
double const EPS = 3e-8;
using namespace std;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const int NX = 1e3 + 10;
int isPrime[NX], check[NX];
int sz, prime[NX];
void ini() {
  isPrime[2] = 1;
  prime[sz++] = 2;
  int i, j;
  for (i = 3; i * i < NX; i += 2) {
    if (check[i] == 0) {
      prime[sz++] = i;
      isPrime[i] = 1;
      for (j = i * i; j < NX; j += (i + i)) check[j] = 1;
    }
  }
  while (i < NX) {
    if (check[i] == 0) {
      isPrime[i] = 1;
      prime[sz++] = i;
    }
    i += 2;
  }
}
int n;
set<int> ans;
bool insertkoro(int x) {
  if (isPrime[x]) return 1;
  if (x + x <= n) return 1;
  vector<int> divisor;
  int i;
  for (i = 0; prime[i] * prime[i] <= x; i++) {
    int cnt = 0;
    while (x % prime[i] == 0) {
      cnt++;
      x /= prime[i];
    }
    if (cnt > 0) divisor.push_back(prime[i]);
  }
  if (x > 1) divisor.push_back(x);
  int sz = divisor.size();
  if (sz == 1) return 1;
  for (__typeof((sz)-1) i = (0); i <= (sz)-1; ++i)
    if (ans.find(divisor[i]) == ans.end()) return 1;
  return 0;
}
const int MX = 1e7 + 10;
int yes[MX], inp[MX];
int main() {
  bool ok = 0;
  int n = ({
    int a;
    read(a);
    a;
  }),
      m = ({
        int a;
        read(a);
        a;
      });
  int sum = 0;
  for (__typeof((n)-1) i = (0); i <= (n)-1; ++i) {
    inp[i] = ({
      int a;
      read(a);
      a;
    });
    inp[i] %= m;
    sum += inp[i];
    if (inp[i] == 0) ok = 1;
  }
  if (ok) {
    printf("YES\n");
    return 0;
  }
  yes[0] = 1;
  for (int i = 0; i < n && ok == 0; i++) {
    for (int j = sum; j >= inp[i] && ok == 0; j--) {
      if (yes[j - inp[i]]) {
        if (j % m == 0) ok = 1;
        yes[j] = 1;
      }
    }
  }
  if (ok)
    puts("YES");
  else
    puts("NO");
  return 0;
}
