Educational Codeforces Round 82 B. National Project

Your company was appointed to lay new asphalt on the highway of length nn. You know that every day you can either repair one unit of the highway (lay new asphalt over one unit of the highway) or skip repairing.

Skipping the repair is necessary because of the climate. The climate in your region is periodical: there are gg days when the weather is good and if you lay new asphalt these days it becomes high-quality pavement; after that, the weather during the next bb days is bad, and if you lay new asphalt these days it becomes low-quality pavement; again gg good days, bb bad days and so on.

You can be sure that you start repairing at the start of a good season, in other words, days 1,2,…,g1,2,…,g are good.

You don't really care about the quality of the highway, you just want to make sure that at least half of the highway will have high-quality pavement. For example, if the n=5n=5 then at least 33 units of the highway should have high quality; if n=4n=4 then at least 22 units should have high quality.

What is the minimum number of days is needed to finish the repair of the whole highway?

Input

The first line contains a single integer TT (1≤T≤1041≤T≤104) — the number of test cases.

Next TT lines contain test cases — one per line. Each line contains three integers nn, gg and bb (1≤n,g,b≤1091≤n,g,b≤109) — the length of the highway and the number of good and bad days respectively.

Output

Print TT integers — one per test case. For each test case, print the minimum number of days required to repair the whole highway if at least half of it should have high quality.

Example

Input

 

3
5 1 1
8 10 10
1000000 1 1000000

Output

 

5
8
499999500000
读懂题意多判断几次即可（没有想到更好的解法orz）。因为这有个硬性限制是高质量的路段数必须不少于一半，所以先从这里考虑，求出至少的周期个数，再决定低质量的路段是否需要额外的天数。注意：在高质量路段已经修建完成的情况下，低质量的路段也可以在good days修建。

#include <bits/stdc++.h>
using namespace std;
long long n,g,b;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d%d",&n,&g,&b);
        long long half;
        if(n%2==0)half=n/2;//至少需要完成的高质量的路段数 
        else half=n/2+1;
        long long res=n-half;//剩余的路段数 
        int period=0;//周期个数 
        long long ans=0;//答案 
        if(half%g==0)//如果要完成的高质量路段数能被g整除 
        {
            period=half/g;
            if(res<=b*period)
            {
                ans=(period-1)*(g+b)+g;
                if(res<=(period-1)*b)
                {
                }
                else
                {
                    ans=res+period*g;
                }
            }
            else
            {
                ans=res+period*g;
            }
        }
        else
        {
            period=half/g+1;
            ans+=(period-1)*g+half%g;
            if(res<=(period-1)*b)ans+=(period-1)*b;
            else ans+=res;
        }
        cout<<ans<<endl;
    }
    return 0;
}