1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
//Krzysztof Boryczka
#pragma GCC optimize "O3"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;

#define FOR(i, b, e) for(int i=b; i<=e; i++)
#define FORD(i, b, e) for(int i=b; i>=e; i--)
#define SIZE(x) ((int)x.size())
#define pb push_back
#define st first
#define nd second
#define sp ' '
#define ent '\n'

const int M=61;
const int N=205;

ll dp[N][N][M+2], wdp[N][N][M+2];
ll t[N], pref[N];
ll m;
int n;

ll sum(int a, int b){
	if(a>b) return 0;
	return pref[b]-pref[a-1];
}

int bity(ll k){
	int bits=0;
	while(1ll<<bits <= k) bits++;
	return bits;
}

int drugy(ll val){
	return bity(val^(1ll<<(bity(val)-1)));
}

void solve(){
	cin>>n>>m;
	FOR(i, 1, n) cin>>t[i];
	FOR(i, 1, n) pref[i]=pref[i-1]+t[i];
	FOR(i, 1, n) FOR(j, i+1, n) wdp[i][j][0]=dp[i][j][0]=-INF;
	FOR(k, 1, M) FOR(i, 1, n) FOR(j, i, n){
		dp[i][j][k]=-INF;
		FOR(l, i, j+1) dp[i][j][k]=max(dp[i][j][k], dp[i][l-1][k-1]+dp[l][j][k-1]+sum(l, j));
	}
	FOR(k, 1, M) FOR(i, 1, n) FOR(j, i, n){
		wdp[i][j][k]=-INF;
		ll suff=((1ll<<k)-1)&m;
		if(((1ll<<(k-1))&suff) == 0) continue;
		int drugi=drugy(suff);
		FOR(l, i, j+1) wdp[i][j][k]=max(wdp[i][j][k], dp[i][l-1][k-1]+wdp[l][j][drugi]+sum(l, j));
	}
	cout<<wdp[1][n][bity(m)]<<ent;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	// int tt; cin>>tt;
	// FOR(te, 1, tt)
	solve();
	return 0;
}