 # Duplicate character

### Duplicate Character

Tina is given a string SS which contains the first letter of all the student names in her class. She got a curiosity to check how many people have their names starting from the same alphabet. So given a string SS, she decided to write a code that finds out the count of characters that occur more than once in the string.

### Input format

The first line contains an integer TT, denoting the number of test cases. Each test case consists of a string SS containing only lowercase characters.

### Output format

For each test case on a new line, print the output in the format `character=count` . If multiple characters have more than one count, print all of them separated by space, in alphabetical order. In case no such character is present print −1−1.

### Constraints

1<=T<=71<=T<=7 1<=|S|<=1071<=|S|<=107, where |S||S| denotes length of string SS.

11 ​secon

### Example

#### Input

33 prepbytes java algorithm

e=2 p=2
￼a=2
￼−1

### Sample test case explanation

In the first string character `p` is occuring 2 times and character `e` is occuring 2 times. Printing them in alphabetical order we get,

e=2 p=2

#include <bits/stdc++.h>
using namespace std;
void solveq()
{
string s;
cin>>s;
long int n=s.length();
map<char,int>mp;
map<char,int>::iterator it;
for(int i=0;i<n;i++)
{
mp[s[i]]++;

}
sort(s.begin(),s.end());
int f=0;

for(it=mp.begin();it!=mp.end();it++)
{

`````` if(it->second>1)
{
f=1;
cout<<it->first<<"="<<it->second<<" ";
}
``````

}
if(f==0)
cout<<-1;

}

int main()
{
int t;
cin>>t;
while(t–)
{
solveq();
cout<<endl;
}
}