But who would have thought ,yifenfei After killing some of them , But once again facing the test of the great labyrinth of fate , This is the devil lemon Another organ set up . Need to know , No matter who , If you're trapped in a maze 1 hours , There is no doubt that he will die ！

pitiful yifenfei In order to save them MM, I jumped into the maze without hesitation . Let's help him ！

The great labyrinth of destiny can be seen as a two-dimensional grid array , As shown in the figure below ：

H" title="Problem H">

yifenfei It started in the upper left corner , The goal, of course, is to reach the location of the great devil in the lower right corner . Every grid in the maze is cursed by the goddess of luck or the demon of pain , So each grid has a value , Go there and get the corresponding value automatically .

Now it is stipulated that yifenfei You can only go right or down , You can only go one space at a time . But if you go right , Then you can walk one grid at a time or go to the grid where the number of columns in the row is a multiple of the number of columns in the current row , namely ： If the current grid is （x,y）, The next step could be （x+1,y）,(x,y+1) perhaps (x,y*k)

among k>1.

In order to be able to eliminate the devil to the greatest extent lemon,yifenfei Hope to be able to get the maximum lucky value in this big maze of fate .

H" title="Problem H">

The first line of each set of test data is two integers n,m, It represents the number of rows and columns respectively (1<=n<=20,10<=m<=1000);

Next is n Row data , Each row contains m It's an integer , Express n That's ok m The lucky value corresponding to the lattice of the column K ( |k|<100 ).

10 -10 10 10

2 11 10 -20

11 2 10 -10 -10

k=2;k<=j;k++)

if(j%k==0)

d=max(d,dp[i][j/k]);// Compared with those who take a few steps, the biggest

#include

#include

#define M 1500

#define N 30

#define INT 0x3f3f3f3f

using namespace std;

int t,n,m,dp[N][M];

int main()

{

//freopen("in.txt","r",stdin);

scanf("%d",&t);

while(t--)

{

scanf("%d%d",&n,&m);

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

scanf("%d",&dp[i][j]);

for(int i=0;i<=n;i++)

{

dp[i][0]=-INT;

dp[0][i]=-INT;

}

dp[0][1]=dp[1][0]=0;

for(int i=1;i<=n;i++)

for(int j=1;j<=m;j++)

{

if(i==1&&j==1)// The first step is not state transition

dp[i][j]=dp[i][j];

else if(j==1)// The one on the far left can't be transferred to the right

dp[i][j]+=dp[i-1][j];

else

{

int d=dp[i][j-1];// It's a step to the right

for(int k=2;k<=j;k++)

if(j%k==0)

d=max(d,dp[i][j/k]);// Compared with those who take a few steps, the biggest

if(i==1)

dp[i][j]+=d;

else

dp[i][j]+=max(d,dp[i-1][j]);

}

}

//for(int i=1;i<=n;i++)

//{

// for(int

j=1;j<=m;j++)

//

printf("%d ",dp[i][j]);

//

printf("\n");

//}

//printf("\n");

printf("%d\n",dp[n][m]);

}

return

0;

}

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