# PinkAndBlue

ptrdung
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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;
#define pii pair<int,int>
#define piii pair<pii,int>
const int mn =1e5+6;
const int maxS = 2e6+6;

template <typename T>
class Vector{
private:
int size;
int capacity;
T *array;
void expand(int newCapacity)
{
if (newCapacity <= size)
return;

T * old = array;
array = new T[newCapacity];

for (int i = 0; i < size; i++)
array[i] = old[i];

delete[] old;
capacity = newCapacity;
}
public:
Vector(int initCapacity = 1)
{
size = 0;
capacity = initCapacity;
array = new T[capacity];
}
~Vector()
{
delete[] array;
}
Vector & operator=(Vector & rhs)
{
if (this != &rhs)
{
delete[] array;
size = rhs.size;
capacity = rhs.capacity;
array = new T[capacity];
for (int i = 0; i < size; i++)
array[i] = rhs.array[i];
}

return *this;
}
int Size()
{
return size;
}
T & operator[](int index)
{
return array[index];
}
void push_back(T newElement)
{
if (size == capacity)
expand(2 * size);
array[size] = newElement;
size++;
}
void popBack()
{
size--;
}
void clear()
{
size = 0;
}
};
Vector<int> g[mn];
char a[mn];
int no[mn],dx[mn],dy[mn],xet[mn];
int n,m,dem1,dem2,res,check;
void dfs(int x)
{
for(int i=0;i<g[x].Size();i++)
{
int y=g[x][i];
if(xet[y]==xet[x]){check=1;break;}
if(xet[y]!=0)continue;
if(check==0)
{
xet[y]=3-xet[x];
if(xet[y]==1&&a[y]=='G')dem1++;
if(xet[y]==2&&a[y]=='B')dem1++;
if(xet[y]==1&&a[y]=='B')dem2++;
if(xet[y]==2&&a[y]=='G')dem2++;
dfs(y);
}
}
}
int main()
{
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//freopen("1.txt","r",stdin);
//freopen(".out","w",stdout);
int te=10;
cin>>te;
for(int tes=1;tes<=te;tes++)
{
//cout<<"Case "<<tes<<"\n";
res=0;
check=0;
cin>>n>>m;
for(int i=1;i<=n;i++){
cin>>a[i];
g[i].clear();
xet[i]=0;
}
for(int i=1;i<=m;i++)
{
int x,y;
cin>>x>>y;
g[x].push_back(y);
g[y].push_back(x);
}
for(int i=1;i<=n;i++)
if(xet[i]==0)
{
dem1=dem2=0;
xet[i]=1;
if(xet[i]==1&&a[i]=='G')dem1++;
if(xet[i]==2&&a[i]=='B')dem1++;
if(xet[i]==1&&a[i]=='B')dem2++;
if(xet[i]==2&&a[i]=='G')dem2++;
dfs(i);
if(dem1<dem2)res+=dem1;
else res+=dem2;
}
//for(int i=1;i<=n;i++)cout<<xet[i]<<" ";
if(check)cout<<-1<<'\n';
else cout<<res<<'\n';
}
return 0;
}
```