/**
          题意：给出一个矩阵，然后求矩阵中有多少个‘##’（上下左右）都行
          做法：二分图最大匹配,匈牙利算法，时间复杂度是O(VE),
**/
#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#define maxn 670
using namespace std;
char ch[maxn][maxn];
int mmap[maxn][maxn];
int g[maxn][maxn];
int linker[maxn];
int used[maxn];
int num = 0;
int n;
int  dfs(int u)
{
          for(int v = 0;v<num;v++)
          {
                    if(g[u][v] && used[v] == 0)
                    {
                              used[v] = 1;
                              if(linker[v] == -1 || dfs(linker[v]))
                              {
                                        linker[v] = u;
                                        return 1;
                              }
                    }
          }
          return 0;
}
int hungary()
{
          memset(linker,-1,sizeof(linker));
          int res = 0;
          for(int i=0;i<num;i++)
          {
                    memset(used,0,sizeof(used));
                    if(dfs(i)) res++;
          }
          return res;
}
int main()
{
//#ifndef ONLINE_JUDGE
        // freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
          int T;
          scanf("%d",&T);
          int Case = 1;
          while(T--)
          {
                    scanf("%d",&n);
                    num = 0;
                    for(int i=0;i<n;i++)
                    {
                              scanf("%s",ch[i]);
                    }
                    for(int i=0;i<n;i++)
                    {
                              for(int j=0;j<n;j++)
                              {
                                        if(ch[i][j] == ‘#‘) mmap[i][j] = num++;
                              }
                    }
                    memset(g,0,sizeof(g));
                    for(int i=0;i<n;i++)
                    {
                              for(int j=0;j<n;j++)
                              {
                                        if(ch[i][j] == ‘#‘)
                                        {
                                                  if(i<n-1 && ch[i+1][j] == ‘#‘)
                                                  {
                                                            g[mmap[i][j]][mmap[i+1][j]] = 1;
                                                            g[mmap[i+1][j]][mmap[i][j]] = 1;
                                                  }
                                                  if(j<n-1 && ch[i][j+1] == ‘#‘)
                                                  {
                                                            g[mmap[i][j]][mmap[i][j+1]] = 1;
                                                            g[mmap[i][j+1]][mmap[i][j]] = 1;
                                                  }
                                        }
                              }
                    }
                   int re = hungary();
                    printf("Case %d: %d\n",Case++,re/2);
          }
          return 0;
}