A. Anti Light‘s Cell Guessing

You are playing a game on a n×mn×m grid, in which the computer has selected some cell (x,y)(x,y) of the grid, and you have to determine which one.

To do so, you will choose some kk and some kk cells (x1,y1),(x2,y2),…,(xk,yk)(x1,y1),(x2,y2),…,(xk,yk), and give them to the computer. In response, you will get kk numbers b1,b2,…bkb1,b2,…bk, where bibi is the manhattan distance from (xi,yi)(xi,yi) to the hidden cell (x,y)(x,y) (so you know which distance corresponds to which of kk input cells).

After receiving these b1,b2,…,bkb1,b2,…,bk, you have to be able to determine the hidden cell. What is the smallest kk for which is it possible to always guess the hidden cell correctly, no matter what cell computer chooses?

As a reminder, the manhattan distance between cells (a1,b1)(a1,b1) and (a2,b2)(a2,b2) is equal to |a1−a2|+|b1−b2||a1−a2|+|b1−b2|.

Input

The first line of the input contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. The description of test cases follows.

The single line of each test case contains two integers nn and mm (1≤n,m≤1091≤n,m≤109) — the number of rows and the number of columns in the grid.

Output

For each test case print a single integer — the minimum kk for that test case.

Example

input

Copy

2
2 3
3 1

output

Copy

2
1

思路：本题大意是一个点可以最少通过到几个点的曼哈顿距离确定这个点（如下图）

n=1&&m=1时不需要其他点，n=1 || m = 1时需要一个点

其他情况只需要两个点

#include<iostream>
#include<cstring>
#include<algorithm>using namespace std;const int N = 1e9 + 10;
int t, n, m;int main()
{cin >> t;while(t --){cin >> n >> m;if(n == 1 && m == 1) cout <<"0" << endl;else if (n == 1 || m == 1) cout << "1" << endl;else cout << "2" << endl;}return 0;
}

（萌新上路，有更好的方法或者写的有问题可以多多交流）

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