You are given two integers a and b, and q queries. The i-th query consists of two numbers l_i and r_i, and the answer to it is the number of integers x such that l_i ≤ x ≤ r_i, and ((x mod a) mod b) ≠ ((x mod b) mod a). Calculate the answer for each query.

Recall that y mod z is the remainder of the division of y by z. For example, 5 mod 3 = 2, 7 mod 8 = 7, 9 mod 4 = 1, 9 mod 9 = 0.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then the test cases follow.

The first line of each test case contains three integers a, b and q (1 ≤ a, b ≤ 200; 1 ≤ q ≤ 500).

Then q lines follow, each containing two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}) for the corresponding query.

Output

For each test case, print q integers — the answers to the queries of this test case in the order they appear.

Example

Input


2
4 6 5
1 1
1 3
1 5
1 7
1 9
7 10 2
7 8
100 200


Output


0 0 0 2 4 
0 91 
#include <bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define int long long
using namespace std;

void solve()
{
    int a,b,q;
    cin >> a >> b >> q;
    vector <int> d(0);
    int lcm = a * b / (__gcd(a, b));
    for (int i = 0; i < lcm; i++)
    {
        if (i % a % b == i % b % a)
            d.push_back(i);
    }
    while (q--)
    {
        int l,r,ans = 0;
        cin >> l >> r;
        ans = (r - l + 1);
        for (int i : d)
        {
            if (i <= r) ans -= (r - i - 1) / lcm + 1;
            if (i <= l - 1) ans += (l - i - 2) / lcm + 1;
        }
        cout << ans << " ";
    }
    cout << "\n";
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("file.inp","r",stdin);
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}
