There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see.

If two trees and Sherlock are in one line, Farmer Sherlock can only see the tree nearest to him.
Input
The first line contains one integer t, represents the number of test cases. Then there are multiple test cases. For each test case there is one line containing two integers m and n(1 ≤ m, n ≤ 100000)
Output

思路：那麼我們可以找下規律，他能看到的一定是x，y座標互素的點，因為如果不互素，那麼就說明他是某個點的倍數，那麼他一定是被這個點擋住了，看不見。所以我們就可以求m行裡面與n互素的數，每一行都做一次容斥，加和就可以了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define LL long long
using namespace std;
const int maxn = 1e6 + 1e5;
LL que[maxn], factor[maxn], num;

void Divid(LL n)
{
    num = 0;
    for(LL i = 2; i * i <= n; ++i)
    {
        if(n % i == 0)
        {
            while(n % i == 0)
                n /= i;
            factor[num++] = i;
        }
    }
    if(n != 1)
        factor[num++] = n;
}

LL solve(LL n)
{
    LL k, t, ans;
    t = ans = 0;
    que[t++] = -1;
    for(LL i = 0; i < num; ++i)	//外迴圈遍歷所有的因子
    {
        k = t;
        for(LL j = 0; j < k; ++j)	//我們這裡每次都是基於前面的乘積，所以這樣下來就把所有情況都遍歷完了
            que[t++] = -1 * que[j] * factor[i];
    }
    for(LL i = 1; i < t; ++i)
        ans += n / que[i];
    return ans;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        LL n, m, ans;
        scanf("%lld%lld", &n, &m);
        if(n < m)
            swap(n, m);
        ans = n;
        for(LL i = 2; i <= m; ++i)
        {
            Divid(i);
            ans += (n - solve(n));
        }
        printf("%lld\n",ans);
    }
    return 0;
}