Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. 

Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n). 

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ iy ≤ j x or i x ≤ j y ≤ j x is not allowed). 

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^ 

Input

Each test case will begin with two integers m and n, followed by n integers S 1, S2, S 3 ... S n. 
Process to the end of file. 

Output

Output the maximal summation described above in one line. 

Sample Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

      

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int a[1000010];
int dp[1000010],Max[1000010];

int main()
{
    int n,m,mmax;
    while (~scanf("%d%d",&m,&n))
    {
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        memset(Max,0,sizeof(Max));
        for (int i=1;i<=m;i++)//分成几组
        {
            mmax=-0x7fffffff;
            for (int j=i;j<=n;j++)//j个数分成i组，至少要有i个数
            {
                dp[j]=max(dp[j-1]+a[j],Max[j-1]+a[j]);
                Max[j-1]=mmax;
                mmax=max(mmax,dp[j]);
//                for (int i=0;i<=n;i++)
//                printf ("%d ",dp[i]);
//                printf ("\n");
//                for (int i=1;i<=n;i++)
//                printf ("%d ",Max[i-1]);
//                printf ("\n");
//                printf ("%d\n",mmax);
            }
        }
        printf ("%d\n",mmax);
    }
    return 0;
}