Moderate Modular Mode

YouKn0wWho has two even integers xx and yy. Help him to find an integer nn such that 1≤n≤2⋅10181≤n≤2⋅1018 and nmodx=ymodnnmodx=ymodn. Here, amodbamodb denotes the remainder of aa after division by bb. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Input

The first line contains a single integer tt (1≤t≤1051≤t≤105) — the number of test cases.

The first and only line of each test case contains two integers xx and yy (2≤x,y≤1092≤x,y≤109, both are even).

Output

For each test case, print a single integer nn (1≤n≤2⋅10181≤n≤2⋅1018) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Example

input

Copy

4
4 8
4 2
420 420
69420 42068

output

Copy

Note

In the first test case, 4mod4=8mod4=04mod4=8mod4=0.

In the second test case, 10mod4=2mod10=210mod4=2mod10=2.

In the third test case, 420mod420=420mod420=0420mod420=420mod420=0.

思路：除余时一定要注意前后的大小关系，非常重要！

非常明显除余就是小的数到大的数的距离（差距）

n%x==y%n 说明x到n 的距离与y到n的距离相等，但一定要考虑x，y的大小关系；

x>y时就有x在y的后方，那么y到x的差距就是（x-y）此时要想满足等式，此时的关系就非常微妙了，可以假设n=x+y，y%（x+y)=y,即（x+y）%x=y，=y%x ，要等于y 就有x>y,便满足关系

#include <iostream>
#include <algorithm>using namespace std;//n%x 与y%n 相等，说明相隔 距离相等int main()
{int t;cin>>t;while(t--){int x,y;cin>>x>>y;if(x>y){cout<<(x+y)<<endl;}else{cout<<y-(y%x)/2<<endl;}}return 0;
}

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