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.
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).
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.
3 5 1 2 3
YES
1 6 5
NO
4 6 3 1 1 3
YES
6 6 5 5 5 5 5 5
YES
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.
题目大意：给出一个长度为n的序列，问能否在这个序列中选出一个非空子序列（不一定连续），使得该子序列的所有元素的和能被给出的m整除。
若m<n，求出序列的前缀和模m的值，总共有n个余数，但是余数只有m中可能，根据抽屉原理可得至少有两个前缀和的余数相等，所以选出余数相同的两个前缀和之间的元素就行了。
否则做DP，$f_{i,j}$表示处理了前i个元素，是否有可能余数为j。
话说其实判m和n的关系，直接做DP也没什么关系。。
代码在此。
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