Theater stage is a rectangular field of size n × m. The director gave you the stage's plan which actors will follow. For each cell it is stated in the plan if there would be an actor in this cell or not.
You are to place a spotlight on the stage in some good position. The spotlight will project light in one of the four directions (if you look at the stage from above) — left, right, up or down. Thus, the spotlight's position is a cell it is placed to and a direction it shines.
A position is good if two conditions hold:
- there is no actor in the cell the spotlight is placed to;
- there is at least one actor in the direction the spotlight projects.
Count the number of good positions for placing the spotlight. Two positions of spotlight are considered to be different if the location cells or projection direction differ.
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the plan.
The next n lines contain m integers, 0 or 1 each — the description of the plan. Integer 1, means there will be an actor in the corresponding cell, while 0 means the cell will remain empty. It is guaranteed that there is at least one actor in the plan.
Output
Print one integer — the number of good positions for placing the spotlight.
Examples
input
代码语言：javascript复制2 4
0 1 0 0
1 0 1 0output
代码语言：javascript复制9input
代码语言：javascript复制4 4
0 0 0 0
1 0 0 1
0 1 1 0
0 1 0 0output
代码语言：javascript复制20Note
In the first example the following positions are good:
- the (1, 1) cell and right direction;
- the (1, 1) cell and down direction;
- the (1, 3) cell and left direction;
- the (1, 3) cell and down direction;
- the (1, 4) cell and left direction;
- the (2, 2) cell and left direction;
- the (2, 2) cell and up direction;
- the (2, 2) and right direction;
- the (2, 4) cell and left direction.
Therefore, there are 9 good positions in this example.
求所有0的所有有1的方向共几个。
类似求前缀和，把四个方向都求一下，然后累加就好了。
代码语言：javascript复制#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
int r,c,ans;
int a[1005][1004],s[1005][1004][5];
int main(){
scanf("%d%d",&r,&c);
for(int i=1;i<=r;i )
for(int j=1;j<=c;j ){
scanf("%d",&a[i][j]);
s[i][j][0]=s[i][j-1][0]|a[i][j];
s[i][j][1]=s[i-1][j][1]|a[i][j];
}
for(int i=r;i;i--)
for(int j=c;j;j--){
s[i][j][2]=s[i][j 1][2]|a[i][j];
s[i][j][3]=s[i 1][j][3]|a[i][j];
}
for(int i=1;i<=r;i )
for(int j=1;j<=c;j )
if(a[i][j]==0){
ans =s[i][j][0] s[i][j][1] s[i][j][2] s[i][j][3];
}
printf("%dn",ans);
return 0;
}
