# Question Name:Bishu and Soldiers

``````#include <bits/stdc++.h>
using namespace std;

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n,i;
cin>>n;
vector<int>v(n);
for(i=0;i<n;i++)
cin>>v[i];
sort(v.begin(),v.end());
int q;
cin>>q;
while(q--)
{
int p,sum=0,j=0;
cin>>p;
for(i=0;i<n;i++)
{
if(v[i]>p)
break;
sum+=v[i];
j++;
}
cout<<j<<" "<<sum<<"\n";
}
return 0;
}``````
• Problem Description
Bishu went to fight for Coding Club. There were N soldiers with various powers. There will be Q rounds to fight and in each round Bishu’s power will be varied. With power M, Bishu can kill all the soldiers whose power is less than or equal to M(<=M).

After each round, All the soldiers who are dead in previous round will reborn.Such that in each round there will be N soldiers to fight. As Bishu is weak in mathematics, help him to count the number of soldiers that he can kill in each round and total sum of their powers.

1<=N<=10000

1<=power of each soldier<=100

1<=Q<=10000

1<=power of bishu<=100
• Test Case 1
Input (stdin)4
1 2 3 4
3
3
10
2
Expected Output3 6
4 10
2 3
• Test Case 2
Input (stdin)7
1 2 3 4 5 6 7
3
3
10
2
Expected Output3 6
7 28
2 3 