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

Copy

3
5 1 1
8 10 10
1000000 1 1000000

output

Copy

5
8
499999500000

Note

In the first test case, you can just lay new asphalt each day, since days 1,3,51,3,5 are good.

In the second test case, you can also lay new asphalt each day, since days 11-88 are good.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#define ll              long long
#define PII             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define pb              push_back
#define mk              make_pair
using namespace std;
int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const int mod = 998244353;
const int N = 2e5 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}

ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        ll n, g, b;
        cin >> n >> g >> b;
        ll x = (((n + 2 - 1) / 2)+g-1) / g -1;
        ll rest = max((n + 2 - 1) / 2 - x*g,0ll);
        ll t = rest + x * (g + b);
        cout<<max(t,n)<< endl;
    }
    return 0;
}