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```Pink and Blue
Xenny was a teacher and he had N students. The N children were sitting in a room. Each child was wearing a white T-shirt, with a unique number from the range 1 to N written on it. T-Shirts of pink and blue color were to be distributed among the students by Xenny. This made the students very happy.

Xenny felt that a random distribution of T-Shirts would be very uninteresting. So, he decided to keep an interesting condition:

Every student would get a T-Shirt that is of a different color than his/her friends. That is, if X and Y are friends and X has a Pink T-Shirt, then Y should compulsorily have a Blue T-Shirt, and vice-versa.

Also, Xenny had a belief that Boys should wear blue T-Shirts and Girls should wear pink T-Shirts. If a boy was given a pink T-Shirt or a girl was given a Blue T-Shirt, he called it an inversion.

So, Xenny wanted to distribute T-Shirts in the above-mentioned interesting manner and also wanted to minimize "inversions". Help him solve the task.

Note: There are no disjoint groups of friends in the room. That is, 2 distinct groups with finite number of students do not exist, but exactly 1 group of students exists in the given situation.

Input
The first line is the number of test cases T.

First line of each test case contains 2 space-separated integers - N and M - number of students and number of friendships present respectively.

Second line consists of N space-separated characters, where ith character denotes the gender of the ith student. B: Boy, G: Girl.

M lines follow. Each line consists of 2 space-separated integers, u and v, showing that u is a friend of v and vice-versa.

Output
If Xenny could distribute the T-Shirts in the desired way, print the minimum number of inversions required.
Else, print -1.

Constraints
1 ≤ N ≤ 105
1 ≤ M ≤ 105
1 ≤ u, v ≤ N

Colors of T-Shirt are represented by uppercase characters 'B' and 'G'

Sample

Input

3

3 2

B G B

1 2

1 3

6 9

B B B G G G

3 5

2 6

4 2

6 3

3 1

3 4

6 1

5 1

1 4

6 5

G G G B G G

6 3

1 3

2 3

4 3

5 3

Output

1

-1

2

Explanation

#1

Student 1 can be given a Blue T-Shirt. Hence, Student 2 and 3 would receive Pink T-Shirts. Since, Student 3 is a Boy and has received a Pink T-Shirt, number of inversions = 1.

#include <iostream>

using namespace std;

void Init();
void Solve();

int main(void)
{
// freopen("input.txt", "r", stdin);

int T;
cin >> T;

for (int i = 1; i <= T; i += 1)
{    Init();
Solve();
}

return 0;
}

int N, M;
int ListLen[100006];
int mListLen[100006];
int mColor[100006];
int Color[100006];
int **List;
int Edges[100006][2];

void Init()
{
cin >> N >> M;

for (int i = 1; i <= N; i += 1)
{
ListLen[i] = 0;
mListLen[i] = 0;
mColor[i] = -1;
}

List = new int * [N + 1];

char IN;
for (int i = 1; i <= N; i += 1)
{
cin >> IN;
Color[i] = (IN == 'B') ? 1 : 0;
}

int u, v;
for (int i = 0; i < M; i += 1)
{
cin >> u >> v;
Edges[i][0] = u;
Edges[i][1] = v;

mListLen[u] += 1;
mListLen[v] += 1;
}

for (int i = 1; i <= N; i += 1)
{
List[i] = new int[mListLen[i]];
}

for (int i = 0; i < M; i += 1)
{
u = Edges[i][0];
v = Edges[i][1];

ListLen[u] += 1;
ListLen[v] += 1;

List[u][ListLen[u]] = v;
List[v][ListLen[v]] = u;
}
}

int BFS();

void Solve()
{
mColor[1] = Color[1];

int res1 = BFS();
int res2 = N - res1;

if (res1 > res2)
{
cout << res2 << endl;
}
else
{
cout << res1 << endl;
}

}

int Queue[10000001];
int FrontQ, RearQ;

void PushQ(int);
int PopQ();

bool Visited[100006];

int BFS()
{
for (int i = 1; i <= N; i += 1)
Visited[i] = 0;

FrontQ = RearQ = -1;

PushQ(1);
Visited[1] = 1;

int cnt = 0;

while (FrontQ != RearQ)
{
int u = PopQ();

for (int i = 1; i <= ListLen[u]; i += 1)
{
int v = List[u][i];
if (!Visited[v])
{
Visited[v] = 1;
mColor[v] = 1 - mColor[u];

if (mColor[v] != Color[v])
cnt += 1;

PushQ(v);
}
else
{
if (mColor[v] == mColor[u])
return -1;
}
}
}

return cnt;
}

void PushQ(int x)
{
RearQ += 1;
Queue[RearQ] = x;
}

int PopQ()
{
FrontQ += 1;
return Queue[FrontQ];
}

```