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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 ≤ 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. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S 1, S 2, 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 32 6 -1 4 -2 3 -2 3

Sample Output

68

Hint

Huge input, scanf and dynamic programming is recommended.

题意：

从序列中截取不相交的 m 段连续子段，求最大和

思路：

（1）子问题：第 j 个位置前截取 i 段的最大值

（2）状态：dp[i][j]，表示第 j 个位置前截取 i 段的最大值

（3）状态转移方程：dp[i][j] = max(dp[i][j], dp[i - 1][k] + a[j], dp[i][j - 1] + a[j]);        k 属于[i - 1, j - 1]

令pre[j - 1] = dp[i - 1][k]，状态转移方程变为 dp[j] = max(dp[j - 1], pre[j - 1]) + a[j]

不要忘了剪枝！！！

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 1e9 +7;const int inf = 0x3f3f3f3f;const int N = 1e6 + 10;int dp[N];int pre[N];int a[N];int main(){    int n, m;    while(~scanf("%d%d", &m, &n))    {        for(int i = 1; i <= n; ++i)        {            scanf("%d", &a[i]);        }        memset(dp, 0, sizeof(dp));        memset(pre, 0, sizeof(pre));        int maxx = -inf;        //        dp[i][j] = max(dp[i][j], dp[i - 1][k] + a[j], dp[i][j - 1] + a[j]);        for(int i = 1; i <= m; ++i)         ///m段        {            maxx = -inf;            for(int j = i; j <= n; ++j)     ///在[0, j]内截了i段            {                dp[j] = max(pre[j - 1], dp[j - 1]) + a[j];                pre[j - 1] = maxx;                maxx = max(maxx, dp[j]);            }        }        cout<<maxx<<'\n';    }    return 0;}