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 maxn = 1000101;
long long a[maxn];
const int mod = 1e9 + 7;
const int inf = 1e9;
char L[200], R[200];
map<int, int> m;
long long d[maxn], b[maxn];
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
string getnum(string code) {
  int res = 0;
  int d = 1;
  for (int i = (int)code.length() - 1; i >= 0; i--) {
    res += d * (code[i] - 'A' + 1);
    d *= 26;
  }
  return to_string(res);
}
string ones[] = {"zero",    "one",     "two",       "three",    "four",
                 "five",    "six",     "seven",     "eight",    "nine",
                 "ten",     "eleven",  "twelve",    "thirteen", "fourteen",
                 "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
string tens[]{"twenty", "thirty",  "forty",  "fifty",
              "sixty",  "seventy", "eighty", "ninety"};
bool ispalindrome(string s) {
  int n = s.length();
  for (int i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) return false;
  }
  return true;
}
int was[maxn];
vector<int> ans[maxn];
void printq(int x) { cout << '?' << ' ' << x << endl; }
void printans(int x) { cout << '!' << ' ' << x << endl; }
int sum(int n) {
  int sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long powint(long long x, int a) {
  long long res = 1;
  for (int i = 0; i < a; i++) {
    res *= x;
  }
  return res;
}
long long f(long long x, long long a, long long b, long long c) {
  return b * powint(sum(x), a) + c;
}
struct query {
  int a, b, d;
};
vector<query> v;
bitset<2048> dp;
int main() {
  ios::sync_with_stdio(0);
  int n;
  int m;
  cin >> n;
  cin >> m;
  bool ok = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] % m == 0) ok = 1;
    a[i] = (a[i] % m + m) % m;
  }
  dp[0] = 1;
  for (int i = 0; i < n; i++) {
    dp = (dp << a[i]) | dp;
    dp = (dp >> m) | dp;
  }
  if (dp[m] or ok)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
