JSZKC is going to spend his vacation!

His vacation has NN days. Each day, he can choose a T-shirt to wear. Obviously, he doesn't want to wear a singer color T-shirt since others will consider he has worn one T-shirt all the time.

To avoid this problem, he has MM different T-shirt with different color. If he wears AA color T- shirt this day and BBcolor T-shirt the next day, then he will get the pleasure of f[A][B]f[A][B].(notice: He is able to wear one T-shirt in two continuous days but may get a low pleasure)

Please calculate the max pleasure he can get.

Input Format

The input file contains several test cases, each of them as described below.

• The first line of the input contains two integers N,MN,M (2 \le N \le 100000, 1 \le M \le 100)(2≤N≤100000,1≤M≤100), giving the length of vacation and the T-shirts that JSZKC has.

• The next follows MM lines with each line MM integers. The j^{th}jth integer in the i^{th}ith line means f[i][j]f[i][j] (1\le f[i][j]\le 1000000)(1≤f[i][j]≤1000000).

There are no more than 1010 test cases.

Output Format

One line per case, an integer indicates the answer.

樣例輸入

3 2
0 1
1 0
4 3
1 2 3
1 2 3
1 2 3

樣例輸出

2
9

題目來源

思路： 本題就是讓你求一個n-1的序列。f[a][b]+f[b][c]+f[c][d]+~~~~+  f[x][y]+f[y][z]的最大值。

那麼第0 天（也就是題目中的第一天） 可以穿任意的衣服，那麼第一天就根據第0 天的衣服確定最大價值，我們可以設dp[time][i][j] 表示第0天穿i 第time天穿j 的最大價值，那麼我們就可以得出 dp[a+b][i][j] =max(dp[a][i][k]+dp[b][k][j])。那麼對於該式子我們可以用快速冪處理來降低時間複雜度。

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N =105;

struct node
{
    ll ma[N][N];
};
int n,m;

node mut(node a,node b)
{
    node ans;
    memset(ans.ma,0,sizeof(ans.ma));
    for(int i=1;i<=m;i++){
        for(int j=1;j<=m;j++){
            for(int k=1;k<=m;k++){
                ans.ma[i][j]=max(ans.ma[i][j],a.ma[i][k]+b.ma[k][j]);
            }
        }
    }
    return ans;
}

node quick_pow(node a,int k)
{
    node ans;
    memset(ans.ma,0,sizeof(ans.ma));
    while(k)
    {
        if(k&1) ans=mut(ans,a);
        a=mut(a,a);
        k>>=1;
    }
    return ans;
}

int main()
{
    node tmp;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                scanf("%lld",&tmp.ma[i][j]);
            }
        }

        node ans=quick_pow(tmp,n-1);

        ll maxx=0;
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                maxx=max(maxx,ans.ma[i][j]);
            }
        }
        printf("%lld\n",maxx);
    }

    return 0;
}