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;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
long long bigmod(long long b, long long p, long long m) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    long long temp = bigmod(b, p / 2, m);
    return (temp * temp) % m;
  }
  return (bigmod(b, p - 1, m) * (b % m)) % m;
}
char inp[10010];
string scan() {
  scanf("%s", inp);
  string s;
  for (int i = 0; inp[i]; i++) s += inp[i];
  return s;
}
vector<vector<int> > v;
vector<bool> visited;
int visit[100];
void dfs(int node) {
  visit[node] = 1;
  for (int i = 0; i < (int)v[node].size(); i++)
    if (!visit[v[node][i]]) dfs(v[node][i]);
}
int connectedcomponenetscnt(int head) {
  int cnt = 0;
  for (int i = 1; i < 500; i++) {
    if (!visit[i]) {
      dfs(i);
      cnt++;
    }
  }
  return cnt;
}
int com(string s, string t) {
  for (int i = 0; i < min(s.length(), t.length()); i++) {
    if (s[i] > t[i]) return 1;
    if (s[i] < t[i]) return 0;
  }
  if (s.length() > t.length()) return 1;
  return 0;
}
int n, m, x;
int a[1001];
int memo[1001][1001][2];
int dp(int i, int sum, int k) {
  if (sum % m == 0 && k) return 1;
  if (i == m) return 0;
  int &ret = memo[i][sum][k];
  if (ret != -1) return ret;
  ret = 0;
  for (int j = 0; j <= a[i]; j++) {
    int H = 0;
    if (j != 0) H = 1;
    ret = ret + dp(i + 1, (sum + i * j) % m, max(k, H));
  }
  return ret;
}
int main() {
  memset(memo, -1, sizeof(memo));
  memset(a, 0, sizeof(a));
  cin >> n >> m;
  int yes = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    x = x % m;
    if (x == 0) yes = 1;
    a[x]++;
    if (a[x] > m) yes = 1;
  }
  if (yes)
    cout << "YES";
  else if (dp(1, 0, 0))
    cout << "YES";
  else
    cout << "NO";
}
