// G++ AC  C++  RE
//  挑战程序设计上的代码，我也不太懂
#include<iostream>
#include<stdio.h>
#include<vector>
#include<string.h> 
#include<queue>
using namespace std;
const int dx[4] = { -1, 0, 0, 1 }, dy[4] = { 0, -1, 1, 0 };
int X, Y;
#define MAX_X 14
#define MAX_Y 14
#define MAX_V 7005
int V;
vector<int>G[MAX_V];
int match[MAX_V];
bool used[MAX_V];
void add_edge(int u, int v)
{
	G[u].push_back(v);
	G[v].push_back(u);
}

bool dfs(int v)
{
	used[v] = true;
	for (int i = 0; i<G[v].size(); i++)
	{
		int u = G[v][i], w = match[u];
		if (w<0 || !used[w] && dfs(w))
		{
			match[v] = u;
			match[u] = v;
			return true;
		}
	}
	return false;
}



char field[MAX_X][MAX_Y];
vector<int>dX, dY;
vector<int>pX, pY;

int dist[MAX_X][MAX_Y][MAX_X][MAX_Y];


void bfs(int x, int y, int d[MAX_X][MAX_Y])
{
	queue<int>qx, qy;
	d[x][y] = 0;
	qx.push(x);
	qy.push(y);

	while (!qx.empty())
	{
		x = qx.front(); qx.pop();
		y = qy.front(); qy.pop();
		for (int k = 0; k<4; k++)
		{
			int x2 = x + dx[k], y2 = y + dy[k];
			if (0 <= x2&&x2<X && 0 <= y2&&y2<Y&&field[x2][y2] == '.'&&d[x2][y2]<0)
			{
				d[x2][y2] = d[x][y] + 1;
				qx.push(x2);
				qy.push(y2);
			}
		}
	}
}

void solve()
{
	int n = X*Y;
		dX.clear(); dY.clear();
		pX.clear(); pY.clear();
		memset(dist, -1, sizeof(dist));

		for (int x = 0; x<X; x++)
		{
			for (int y = 0; y<Y; y++)
			{
				if (field[x][y] == 'D')
				{
					dX.push_back(x);
					dY.push_back(y);

					bfs(x, y, dist[x][y]);
				}
				else if (field[x][y] == '.')
				{
					pX.push_back(x);
					pY.push_back(y);
				}
			}
		}
		int d = dX.size(), p = pX.size();
		for (int i = 0; i<d; i++)
		{
			for (int j = 0; j<p; j++)
			{
				if (dist[dX[i]][dY[i]][pX[j]][pY[j]] >= 0)
				for (int k = dist[dX[i]][dY[i]][pX[j]][pY[j]]; k <= n; k++)
				{
					add_edge((k - 1)*d + i, n*d + j);
				}
			}
		}

		if (p == 0)
		{
			puts("0");
			return;
		}
		int num = 0;
		memset(match, -1, sizeof(match));
		for (int v = 0; v<n*d; v++)
		{
			memset(used, 0, sizeof(used));
			if (dfs(v))
			{
				if (++num == p)
				{
					printf("%d\n", v / d + 1);
					return;
				}
			}
		
		}		
		   printf("impossible\n");
		   return;

}

int main()
{
	//freopen("c:in.txt", "r", stdin);
	int T;
	cin >> T;
	while (T--)
	{
		cin >> X >> Y;
		getchar();
		for (int x = 0; x<X; x++)
		{
			for (int y = 0; y<Y; y++)
			{
				scanf("%c", &field[x][y]);
			}
			getchar();
		}
		solve();
		for(int i=0;i<20001;i++)
		G[i].clear();
	}

}