Educational Codeforces Round 32 E

E. Maximum Subsequence

You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequence of indices b₁, b₂, ..., b_k (1 ≤ b₁ < b₂ < ... < b_k ≤ n) in such a way that the value of is maximized. Chosen sequence can be empty.

Print the maximum possible value of .

Input

The first line contains two integers n and m (1 ≤ n ≤ 35, 1 ≤ m ≤ 10⁹).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print the maximum possible value of .

Examples

Input

4 4
5 2 4 1

Output

3

Input

3 20
199 41 299

Output

19

Note

In the first example you can choose a sequence b = {1, 2}, so the sum is equal to 7 (and that's 3 after taking it modulo 4).

In the second example you can choose a sequence b = {3}.

有一串数列   可以从中任取k个数字   令k个数字相加除摸m   求最大的值。

n<=35  如果直接dfs的话  是2^35看到会爆掉，所以可以分两次  这样最大也就是2^18  才二十多万看到能过了。

先求出前半部分的，在求出后半部分的。然后在其中一个里找某个数  使的更接近m就行，这样就要用到二分了

总的复杂度是2^(n/2)*log(n/2)  足够算出这个题目了。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1<<18;
ll a[40], b[40], c[N], d[N], k, ans, sum, n, m;
void dfs(int x) {
    if(x == ans + 1) {
        c[++k] = sum%m;
        return ;
    }
    sum += b[x]; dfs(x + 1);
    sum -= b[x]; dfs(x + 1);
}
int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> a[i], a[i] %= m, b[i] = a[i];
    ans = n/2;
    dfs(1);
    ll n1 = k;
    for(int i = 1; i <= k; i ++) d[i] = c[i];
    memset(c, 0, sizeof(c));
    ans = n; k = 0;
    dfs(n/2+1);
    ll MAX = -1;
    sort(d+1,d+n1+1);
    for(int i = 1; i <= k; i ++) {
        int t = lower_bound(d+1,d+n1,m-c[i]-1)-d;  
        while(c[i]+d[t]>=m)t--;  
        MAX=max(MAX,c[i] + d[t]);  
    }
    printf("%lld
",MAX);
    return 0;
}