#include<stdio.h>
#include<string.h>
#include<queue>using namespace std;const int maxn = 50+5;struct node
{int x;int y;int step;
};
int N,ei,ej,si,sj;
int vis[maxn][maxn];
int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};    //记录对应编号的方向
int mp[maxn][maxn];void init()
{memset(mp,0,sizeof(mp));memset(vis,0,sizeof(vis));
}
bool ok(int x,int y,int k)
{if(x >= N || y >= N || x < 0 || y < 0) return false;//流出边界了int di = 1<<k;if(!(mp[x][y]&di)) return false;    //这里！优先级大于&，所以要！(...)return true;
}int bfs(int x,int y)
{struct node cur,next;queue<node> q;int ans = 0;cur.x = x;cur.y = y;cur.step = 0;q.push(cur);vis[x][y] = 1;int mstep = 1e9+5;//相当于取无穷while(!q.empty()){cur = q.front();q.pop();if(cur.step >= mstep) break;//如果在这一步已经有其他节点流出或到终点了，就不用拓展它的子节点了（因为流出的那一步的子节点肯定不流过了）for(int i = 0;i < 4;i++){if((1<<i)&mp[cur.x][cur.y]){next.x = cur.x+dir[i][0];next.y = cur.y+dir[i][1];if((next.x == ei && next.y == ej) || !ok(next.x,next.y,(i+2)%4))    //可以手算一下感受为什么是(i+2)%4;{mstep = cur.step+1;     //因为流出或到终点了，记录流出的是第几步continue;}if(!vis[next.x][next.y])        //这些节点都是成立的{vis[next.x][next.y] = 1;next.step = cur.step+1;q.push(next);ans++;}}}}return ans;
}
int main()
{//freopen("in.txt","r",stdin);int T;scanf("%d",&T);while(T--){init();scanf("%d",&N);for(int i = 0;i < N;i++){for(int j = 0;j < N;j++){scanf("%d",&mp[i][j]);if(mp[i][j]&32){ei = i;ej=j;mp[i][j] -= 32;}else if(mp[i][j]&16){si = i;sj = j;mp[i][j] -= 16;}//printf("%d ",mp[i][j]);}//printf("\n");}printf("%d\n",bfs(si,sj));}return 0;
}