Jewels and Stones

Problem Statement

Soma is a fashionable girl. She absolutely loves shiny stones that she can put on as jewellery accessories. She has been collecting stones since her childhood - now she has become really good with identifying which ones are fake and which ones are not. Her King requested for her help in mining precious stones, so she has told him which all stones are jewels and which are not. Given her description, your task is to count the number of jewel stones. 

More formally, you're given a string J composed of latin characters where each character is a jewel. You're also given a string S composed of latin characters where each character is a mined stone. You have to find out how many characters of S are in J as well.

Input

First line contains an integer T denoting the number of test cases. Then follow T test cases. Each test case consists of two lines, each of which contains a string composed of English lower case and upper characters. First of these is the jewel string J and the second one is stone string S. 
You can assume that 1 <= T <= 100, 1 <= |J|, |S| <= 100

Output

Output for each test case, a single integer, the number of jewels mined.

Example

Input:
4
abc
abcdef
aA
abAZ
aaa
a
what
none

Output:
3
2
1
0

Problem link: stones

The idea: The idea is to find the no of characters in 2nd string   which are already present in the first string. One can 

use the  stl map to record the characters already present in the 1st string.

The code:

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        string first;
        string second;
        cin>>first>>second;
        map<char,bool> mapf;
        for(int i=0;i<first.length();i++)
        {
            mapf[first[i]]=true;
        }
        int count=0;
        for(int i=0;i<second.length();i++)
        {
            if(mapf[second[i]])
            count++;
        }
        cout<<count<<endl;
        
        
    }
}