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 S₁, S₂, S₃, S₄ ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + sum(i_m, j_m) maximal (i_x ≤ i_y ≤ 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. ^_^

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

Output the maximal summation described above in one line.

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

6
8

Hint
Huge input, scanf and dynamic programming is recommended.
 
 #include <iostream>
using namespace std;
#define N 1000005
#define INIT -99999999
#define max(a,b) a>b?a:b

int pre[N];    //记录MAX出现的值
int now[N];    //储存当前位置内的最大值
int val[N];    //储存接收的值
int MAX;

int main()
{
	int n,m,i,j;
	while (scanf("%d",&m)!=EOF)
	{
		MAX=INIT;
		memset(pre,0,sizeof(pre));
		memset(now,0,sizeof(now));
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%d",&val[i]);
		for(j=1;j<=m;j++)
		{
			MAX=INIT;      //前面的j-1个值给前面的j-1个串
			pre[j-1]=MAX;
			for(i=j;i<=n;i++)
			{
				now[i]=max(pre[i-1]+val[i],now[i-1]+val[i]);
				pre[i-1]=MAX;
				MAX=max(MAX,now[i]);
			}	
		}
		printf("%d/n",MAX);
	}
	return 0;
}