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.

def gcd(a, b): return a if b == 0 else gcd(b, a % b)


def good(a, m):
    if len(a) > m:
        return True
    b = [0] * m
    for e in a:
        b[e % m] += 1
    if b[0] > 0:
        return True
    for r in range(1, (m + 1) // 2):
        if b[r] > 0 and b[m - r] > 0:
            return True
    for r in range(m):
        p = m // gcd(r, m)
        if b[r] >= p:
            return True
    if m % 2 == 0 and b[m // 2] > 1:
        return True
    d = 2
    while d * d <= m:
        if m % d == 0:
            if b[d] >= m // d:
                return True
            if b[m // d] >= d:
                return True
        d += 1
    if b[1] >= m or b[m - 1] >= m:
        return True
    can = [[False] * m for _ in range(len(a) + 1)]
    for n in range(1, len(a) + 1):
        can[n][a[n - 1] % m] = True
        for r in range(m):
            can[n][r] = can[n][r] or can[n - 1][r] or can[n - 1][(r - a[n - 1]) % m]
    return can[len(a)][0]
    # mx = []
    # for r in range(m):
    #     seen = [False] * m
    #     i = 1
    #     while not seen[r * i % m]:
    #         seen[r * i % m] = True
    #         i += 1
    #     mx.append(i)
    # dp0 = [[False] * m for _ in range(m + 1)]
    # dp1 = [[False] * m for _ in range(m + 1)]
    # dp0[0][0] = True
    # for n in range(1, m + 1):
    #     for r in range(m):
    #         dp0[n][r] = dp0[n - 1][r]
    #         dp1[n][r] = dp1[n - 1][r]
    #         for c in range(1, min(mx[n - 1], b[n - 1] + 1)):
    #             dp0[n][r] = dp0[n][r] or dp0[n - 1][(r - (n - 1) * c) % m]
    #             dp1[n][r] = dp1[n][r] or dp0[n - 1][(r - (n - 1) * c) % m]
    # return dp1[m][0]


n, m = map(int, input().split())
a = list(map(int, input().split()))
print("YES" if good(a, m) else "NO")