B. Spotlights

                                                                                                             B. Spotlights

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 andm (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, while0 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

2 40 1 0 01 0 1 0

Output

9

Input

4 40 0 0 01 0 0 10 1 1 00 1 0 0

Output

20

Note

In the first example the following positions are good:

1. the (1, 1) cell and right direction;
2. the (1, 1) cell and down direction;
3. the (1, 3) cell and left direction;
4. the (1, 3) cell and down direction;
5. the (1, 4) cell and left direction;
6. the (2, 2) cell and left direction;
7. the (2, 2) cell and up direction;
8. the (2, 2) and right direction;
9. the (2, 4) cell and left direction.

Therefore, there are 9 good positions in this example.

题意：给出一个矩阵，对于每个0统计其上下左右有几个方向是1，该点的对应值就是几。计算矩阵中所有0的对应值之和。其实看样例就可以懂。

题解1；还可以把每行的第一个1和最后一个1标记一下，0在这两个1中间的就是+2，不在就+1.对于每一行每一列都是这样的，复杂度也是如下

题解2：先跑一次纵向，求出每个点在纵向上的对应值之和。再跑一次横向，求出每个点在横向上的对应值之和。总体复杂度为 O(2*n^2)

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace  std;#define  maxn 1000+10typedef long long LL;int a[maxn][maxn];int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){for(int i=0;i<n;i++)for(int j=0;j<m;j++)scanf("%d",&a[i][j]);int ans=0;for(int i=0;i<n;i++){int cnt=0,flag=0;for(int j=0;j<m;j++){if(a[i][j]==0&&(!flag))cnt++;else if(a[i][j]==1){ans+=cnt;cnt=0;flag=1;}else if(a[i][j]==0&&flag){ans++;cnt++;}}}for(int j=0;j<m;++j){int cnt=0,flag=0;for(int i=0;i<n;++i){if(a[i][j]==0&&(!flag))cnt++;else if(a[i][j]==1){ans+=cnt;cnt=0;flag=1;}else if(a[i][j]==0&&flag){ans++;cnt++;}}}printf("%d\n",ans);}return 0;}