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, ..., bk (1 ≤ b1 < b2 < ... < bk ≤ 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, ..., an (1 ≤ ai ≤ 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}.
题意：
给你一个数组a，和一个数m，让你选出一个数组a的子集，让子集中的数的sum和对m取模后的数值最大。
思路：
meet-in-the-middle 的思路，把数组分成两个部分，每一个部分数最多就是 35/2 的大小，对于每一个部分，我们通过二进制枚举所有状态， 把第一部分的所有子集的sum和求出（同意取模），然后第二部分同样的操作，把取模后的sum分别加入到数组v1和v2中，最后对于第一部分的每一个数x，去第二部分中找一个数y，使之(x+y )%m 尽量大。
那么x+y 分为两种情况，
当x+y 大于等于m，
那么结果是 x+y-m 的，那么这里的y我们不妨选择最大的，这样可以使数值更大。
反之就是 x+y<m
y要是小于m-x 中的最大值，这样x+y 更接近m-1
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll a[60];
std::vector<ll> v1;
std::vector<ll> v2;
ll haf;
ll n,m;
void dfs1(int id,ll num)
{
v1.pb(num%m);
if(id<=haf)
{
dfs1(id+1,num);
dfs1(id+1,num+a[id]);
}
}
void dfs2(int id,ll num)
{
v2.pb(num%m);
if(id<=n)
{
dfs2(id+1,num);
dfs2(id+1,num+a[id]);
}
}
int main()
{
// freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
gbtb;
cin>>n>>m;
repd(i,1,n)
{
cin>>a[i];
a[i]%=m;
}
haf=(n>>1);
dfs1(1,0ll);
dfs2(haf+1,0ll);
sort(ALL(v1));
sort(ALL(v2));
v1.erase(unique(ALL(v1)),v1.end());
v2.erase(unique(ALL(v2)),v2.end());
ll ans=0ll;
int p=sz(v2);
int q=sz(v1);
for(auto x:v1)
{
int pos=lower_bound(ALL(v2),m-x)-v2.begin();
if(pos>=1&&pos<p)
{
ans=max(ans,(x+v2[pos-1])%m);
}
ans=max(ans,(x+v2[p-1])%m);
}
cout<<ans<<endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}