A. Pasha and Pixels

A. Pasha and Pixels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.

Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting ofn row with m pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a2 × 2 square consisting of black pixels is formed.

Pasha has made a plan of k moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbersi and j, denoting respectively the row and the column of the pixel to be colored on the current move.

Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the2 × 2 square consisting of black pixels is formed.

Input

The first line of the input contains three integersn, m, k (1 ≤ n, m ≤ 1000,1 ≤ k ≤ 10⁵) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.

The next k lines contain Pasha's moves in the order he makes them. Each line contains two integersi and j (1 ≤ i ≤ n,1 ≤ j ≤ m), representing the row number and column number of the pixel that was painted during a move.

Output

If Pasha loses, print the number of the move when the2 × 2 square consisting of black pixels is formed.

If Pasha doesn't lose, that is, no 2 × 2 square consisting of black pixels is formed during the given k moves, print 0.

Sample test(s)

Input

2 2 41 11 22 12 2

Output

4

Input

2 3 62 32 21 32 21 21 1

Output

5

Input

5 3 72 31 21 14 13 15 33 2

Output

0又一个大水题
#include<stdio.h>#include<string.h>  int d[1001][1001];int main(){    int i,a,b,k;    int x,y;    scanf("%d%d%d",&a,&b,&k);        memset(d,0,sizeof(d));        for(i=1;i<=k;i++)        {            scanf("%d%d",&x,&y);//一个个接受        d[x][y]=1;        if((d[x-1][y]&&d[x][y-1]&&d[x-1][y-1])||(d[x+1][y]&&d[x][y+1]&&d[x+1][y+1])||(d[x+1][y]&&d[x][y-1]&&d[x+1][y-1])||(d[x-1][y]&&d[x][y+1]&&d[x-1][y+1]))           break;//若满足自己四周全为1，就直接跳出，不在接受        }        if(i<=k)        printf("%d\n",i);        else        printf("0\n");    return 0;}