The 17th Zhejiang University Programming Contest Sponsored by TuSimple A

Marjar Cola

---

Time Limit: 1 Second      Memory Limit: 65536 KB

---

Marjar Cola is on sale now! In order to attract more customers, Edward, the boss of Marjar Company, decides to launch a promotion: If a customer returns x empty cola bottles or y cola bottle caps to the company, he can get a full bottle of Marjar Cola for free!

Now, Alice has a empty cola bottles and b cola bottle caps, and she wants to drink as many bottles of cola as possible. Do you know how many full bottles of Marjar Cola she can drink?

Note that a bottle of cola consists of one cola bottle and one bottle cap.

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 100), indicating the number of test cases. For each test case:

The first and only line contains four integers x, y, a, b (1 ≤ x, y, a, b ≤ 100). Their meanings are described above.

Output

For each test case, print one line containing one integer, indicating the number of bottles of cola Alice can drink. If Alice can drink an infinite number of bottles of cola, print "INF" (without the quotes) instead.

Sample Input

2
1 3 1 1
4 3 6 4

Sample Output

INF
4

Hint

For the second test case, Alice has 6 empty bottles and 4 bottle caps in hand. She can return 4 bottles and 3 caps to the company to get 2 full bottles of cola. Then she will have 4 empty bottles and 3 caps in hand. She can return them to the company again and get another 2 full bottles of cola. This time she has 2 bottles and 2 caps in hand, but they are not enough to make the exchange. So the answer is 4.

题意：换一瓶饮料需要a或者b种材料，现在我有x,y个材料，问最多换几个，如果是无限就输出INF

解法：模拟，设置一个上限，如果超过就输出INF

#include<bits/stdc++.h>
using namespace std;
long long t;
long long a,b,c,d;
int main()
{
    while(cin>>t)
    {
        while(t--)
        {
            cin>>a>>b>>c>>d;
            long long an,bn;
            long long sum=0;
            long long x=0;
            an=c;
            bn=d;
            long long y=0;
            while(an>=a||bn>=b)
            {
             //   cout<<an<<" "<<bn<<endl;
                sum+=(an/a+bn/b);
                long long an1=(an/a+bn/b)+an%a;
                long long bn1=(an/a+bn/b)+bn%b;
                if(an1>=an&&bn1>=bn)
                {
                    x=1;
                    break;
                }
                else
                {
                    an=an1;
                    bn=bn1;
                }
                y++;
                if(y>=100)
                {
                    x=1;
                    break;
                }
            }
            if(x)
            {
                cout<<"INF"<<endl;
            }
            else
            {
                cout<<sum<<endl;
            }
        }
    }
    return 0;
}