Problem:
Score : 800 points

Problem Statement

There is a pond with a rectangular shape.
The pond is divided into a grid with H rows and W columns of squares.
We will denote the square at the i-th row from the top and j-th column from the left by (i,\ j).

Some of the squares in the pond contains a lotus leaf floating on the water.
On one of those leaves, S, there is a frog trying to get to another leaf T.
The state of square (i,\ j) is given to you by a character a_{ij}, as follows:

. : A square without a leaf.

o : A square with a leaf floating on the water.

S : A square with the leaf S.

T : A square with the leaf T.

The frog will repeatedly perform the following action to get to the leaf T: "jump to a leaf that is in the same row or the same column as the leaf where the frog is currently located."

Snuke is trying to remove some of the leaves, other than S and T, so that the frog cannot get to the leaf T.
Determine whether this objective is achievable. If it is achievable, find the minimum necessary number of leaves to remove.

Constraints

2 ≤ H, W ≤ 100

a_{ij} is ., o, S or T.

There is exactly one S among a_{ij}.

There is exactly one T among a_{ij}.

Input

Input is given from Standard Input in the following format:

H W
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}

Output

If the objective is achievable, print the minimum necessary number of leaves to remove.
Otherwise, print -1 instead.

Sample Input 1

3 3
S.o
.o.
o.T

Sample Output 1

2

Remove the upper-right and lower-left leaves.

Sample Input 2

3 4
S...
.oo.
...T

Sample Output 2

0

Sample Input 3

4 3
.S.
.o.
.o.
.T.

Sample Output 3

-1

Sample Input 4

10 10
.o...o..o.
....o.....
....oo.oo.
..oooo..o.
....oo....
..o..o....
o..o....So
o....T....
....o.....
........oo

Sample Output 4

5

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#ARC074-F Lotus Leaves
"""
この問題は最小カットに帰着できる。但し、h*wの頂点を持つと間に合わないので、H軸とW軸を独立に考える。
縦H個、横W個の頂点を持つ。
ある座標h,wがoの時、頂点H[h]とW[w]と容量1のネットワークでつなぐ。[sourse,h0,h1,...hH,w0,w1,...wW,sink]
startとgoalを最大流の始点、終点として持つのではなく、それとは別にsourse,sinkの2頂点を作る。
startとsourse,goalとsinkを容量INFでつなぐことにより、より簡単に最小カットを求めることが可能になる。
よって、グラフ長はH*W+2となる。
ネットワーク自体は無向なので、undirectedを使用する。
"""
import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0

from collections import deque

class Dinic:
    def __init__(self, N, source, sink):
        self.N = N
        self.G = [[] for _ in range(N)]
        self.source = source
        self.sink = sink
 
    def add_edge(self, fr, to, cap):
        n1 = len(self.G[fr])
        n2 = len(self.G[to])
        self.G[fr].append([to, cap, n2])
        self.G[to].append([fr, 0, n1]) # 逆辺を cap 0 で追加
        
    def add_edge_undirected(self, fr, to, cap):
        n1 = len(self.G[fr])
        n2 = len(self.G[to])
        self.G[fr].append([to, cap, n2])
        self.G[to].append([fr, cap, n1]) # 逆辺を cap 0 で追加
        
    def bfs(self):
        level = [0] * self.N
        G = self.G; source = self.source; sink = self.sink
        q = deque([source])
        level[source] = 1
        pop = q.popleft; append = q.append
        while q:
            v = pop()
            lv = level[v] + 1
            for to, cap, rev in G[v]:
                if not cap:
                    continue
                if level[to]:
                    continue
                level[to] = lv
                if to == sink:
                    self.level = level
                    return
                append(to)
        self.level = level
        
    def dfs(self,v,f):
        if v == self.sink:
            return f
        G = self.G
        prog = self.progress
        level = self.level
        lv = level[v]
        E = G[v]
        for i in range(prog[v],len(E)):
            to, cap, rev = E[i]
            prog[v] = i
            if not cap:
                continue
            if level[to] <= lv:
                continue
            x = f if f < cap else cap
            ff = self.dfs(to, x)
            if ff:
                E[i][1] -= ff
                G[to][rev][1] += ff
                return ff
        return 0
    
    def max_flow(self):
        INF = 10**18
        flow = 0
        while True:
            self.bfs()
            if not self.level[self.sink]:
                return flow
            self.progress = [0] * self.N
            while True:
                f = self.dfs(self.source, INF)
                if not f:
                    break
                flow += f
        return flow

H,W = map(int,readline().split())
maze = []
for _ in range(H):
    maze += [list(readline().rstrip().decode())]

source = 0
sink = H+W+1
network = Dinic(H+W+2,source,sink)
INF = 10**18

for i in range(1,H+1):
    for j,ox in enumerate(maze[i-1],1):
        if ox == "x":
            continue
        elif ox == "o":
            network.add_edge_undirected(i,H+j,1)
        elif ox == "S":
            network.add_edge_undirected(source,i,INF)
            network.add_edge_undirected(source,H+j,INF)
        elif ox == "T":
            network.add_edge_undirected(i,sink,INF)
            network.add_edge_undirected(H+j,sink,INF)

f = network.max_flow()
if f == INF:
    print(-1)
else:
    print(f)