Problem Description

A long time ago there are no himalayas between India and China, the both cultures are frequently exchanged and are kept in sync at that time, but eventually himalayas rise up. With that at first the communation started to reduce and eventually died.

Let's assume from my crude drawing that the only way to reaching from India to China or viceversa is through that grid, blue portion is the ocean and people haven't yet invented the ship. and the yellow portion is desert and has ghosts roaming around so people can't travel that way. and the black portions are the location which have mountains and white portions are plateau which are suitable for travelling. moutains are very big to get to the top, height of these mountains is infinite. So if there is mountain between two white portions you can't travel by climbing the mountain.

And at each step people can go to  adjacent positions.

Our archeologists have taken sample of each mountain and estimated at which point they rise up at that place. So given the times at which each mountains rised up you have to tell at which time the communication between India and China got completely cut off.

Input

There are multi test cases. the first line is a sinle integer T which represents the number of test cases.

For each test case, the first line contains two space seperated integers N,M. next N lines consists of strings composed of , characters.  denoting that there's already a mountain at that place, 0 denoting the plateau. on N+2 line there will be an integer Q denoting the number of mountains that rised up in the order of times. Next Q lines contain 2 space seperated integers X,Y denoting that at ith year a mountain rised up at location X,Y.

T≤

≤N≤

≤M≤

≤Q≤N∗M

≤X<N

≤Y<M

Output

Single line at which year the communication got cut off.

print - if these two countries still connected in the end.

Hint:

From the picture above, we can see that China and India have no communication since 4th year.

Sample Input

Sample Output

#include<cstdio>

#include<cstring>

#include<queue>

#include<cstdlib>

#include<algorithm>

#include<iostream>

using namespace std;

#define met(a,b) memset(a,b,sizeof(a));

const int oo = 0x3f3f3f3f;

const int N = ;

char str[N][N],maps[N][N];

int dis[][]={{,},{,},{-,},{,-}};

int x[N*N],y[N*N];

int vis[N][N];

int n,m;

struct node

{

    int x,y;

};

int dfs(int x,int y)

{

    met(vis,);

    queue<node> Q;

    node q,p;

    q.x=x;q.y=y;

    Q.push(q);

    vis[x][y]=;

    while(Q.size())

    {

        q=Q.front();

        Q.pop();

        if(q.x==n-)

            return ;

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

        {

            p.x=q.x+dis[i][];

            p.y=q.y+dis[i][];

            if(p.x>= && p.x<n && p.y>= && p.y<m && !vis[p.x][p.y]&& maps[p.x][p.y]=='')

            {

                vis[p.x][p.y]=;

                Q.push(p);

            }

        }

    }

    return ;

}

int pan()

{

    for(int i=;i<m;i++)

    {

        if(maps[][i]=='')

        {

            if(dfs(,i))

                return ;

        }

    }

    return ;

}

void buile(int mid)

{

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

    {

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

            maps[i][j]=str[i][j];

    }

    for(int i=;i<=mid;i++)

        maps[x[i]][y[i]]='';

}

int main()

{

    int t,q;

    scanf("%d",&t);

    while(t--)

    {

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

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

            scanf("%s",str[i]);

        scanf("%d",&q);

        for(int i=; i<=q; i++)

            scanf("%d %d",&x[i],&y[i]);

        int l=,r=q;

        int mid=;

        while(l<=r)///二分查找查到那一年开始不通的

       {

            met(maps,);///建立新的地图

            mid=(l+r)/;

            buile(mid);

            if(!pan())///如果不通才往前找一年

            {

                r=mid-;

            }

            else        ///否则往后找一年

                l=mid+;

        }

        if(l>q)///如果找到的那年比q大，说明一直是通的输出-1

            l=-;

        printf("%d\n",l);

    }

    return ;

}