2017 JUST Programming Contest 3.0 E. The Architect Omar

E. The Architect Omar

time limit per test

1.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Architect Omar is responsible for furnishing the new apartments after completion of its construction. Omar has a set of living room furniture, a set of kitchen furniture, and a set of bedroom furniture, from different manufacturers.

In order to furnish an apartment, Omar needs a living room furniture, a kitchen furniture, and two bedroom furniture, regardless the manufacturer company.

You are given a list of furniture Omar owns, your task is to find the maximum number of apartments that can be furnished by Omar.

Input

The first line contains an integer T (1 ≤ T ≤ 100), where T is the number of test cases.

The first line of each test case contains an integer n (1 ≤ n ≤ 1000), where n is the number of available furniture from all types. Then nlines follow, each line contains a string s representing the name of a furniture.

Each string s begins with the furniture's type, then followed by the manufacturer's name. The furniture's type can be:

• bed, which means that the furniture's type is bedroom.
• kitchen, which means that the furniture's type is kitchen.
• living, which means that the furniture's type is living room.

All strings are non-empty consisting of lowercase and uppercase English letters, and digits. The length of each of these strings does not exceed 50 characters.

Output

For each test case, print a single integer that represents the maximum number of apartments that can be furnished by Omar

Example

input

16bedXskitchenSS1kitchen2bedXsliving12livingh

output

1

#include <iostream>#include<string>#include<math.h>#include<algorithm>#include<string.h>using namespace std;const int maxn=1010;//int b[maxn],k[maxn],l[maxn];int main(){    int t;    cin>>t;    while(t--)    {        int n;        int b=0,l=0,k=0;        cin>>n;        string str;        while(n--)        {            cin>>str;            if(str[0]=='b')            {                b++;            }            else if(str[0]=='k')            {                k++;            }            else            {                l++;            }        }        int b1=b/2;        int tmp=b1;        tmp=min(tmp,k);        tmp=min(tmp,l);        cout<<tmp<<endl;    }    return 0;}