//双端队列优化
//这道题就相当于dijkstra了，边的权值为0||1
//由于0的特殊性，我们可以使用双端队列来优化堆

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>

using namespace std;
const int N=510;
typedef pair<int,int>PII;


char g[N][N];
int dist[N][N];

int n,m;

int bfs()
{
    memset(dist,0x3f,sizeof dist);
    deque<PII> dq;
    dq.push_front({0,0});
    dist[0][0]=0;
    int dx[4]={-1,-1,1,1},dy[4]={-1,1,1,-1};

    int di[4]={-1,-1,0,0},dj[4]={-1,0,0,-1};
    char ch[]={"\\/\\/"};

    while(dq.size())
    {
        auto op=dq.front();
        dq.pop_front();
        int x=op.first,y=op.second;

        for(int i=0;i<4;i++)
        {
            int wx=x+dx[i],wy=y+dy[i];
            if(wx>=0&&wx<=n&&wy>=0&&wy<=m)
            {
                int w=0;
                int kx=x+di[i],ky=y+dj[i];
                if(g[kx][ky]!=ch[i])w=1;

                if(dist[wx][wy]>dist[x][y]+w)
                {
                    dist[wx][wy]=dist[x][y]+w;

                    if(w)dq.push_back({wx,wy});
                    else dq.push_front({wx,wy});
                }
            }
        }

    }
    if(dist[n][m]==0x3f3f3f3f)return -1;
    return dist[n][m];
}

int main()
{   
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)cin>>g[i];

        int t=bfs();
        if(t==-1)cout<<"NO SOLUTION"<<endl;
        else cout<<t<<endl;
    }
    return 0;
}