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 <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <tuple>
#include <deque>
#include <set>
#include <map>
#include <regex>
#include <unordered_set>
#include <unordered_map>
#include <climits>
#include <iomanip>
#include <math.h>

#define ll long long
#define ld long double
#define INFL LLONG_MAX
#define INFI INT_MAX
#define PI 3.14159265358979323846

using namespace std;



int main() {
    int t;
    cin >> t;
    for (int ti = 1; ti <= t; ti++) {
        ll a, b, q;
        cin >> a >> b >> q;
        ll gcd = __gcd(a, b);
        ll lcm = a * b / gcd;
        ll mn = min(a, b);
        ll mx = max(a, b);
        for (int qi = 0; qi < q; qi++) {
            ll l, r;
            cin >> l >> r;
            ll beforeL = l - (l % lcm);
            ll beforeR = r - (r % lcm);
            ll ans = 0;
            if (beforeL == beforeR) {
                ll currentL = l % lcm;
                ll currentR = min(mx - 1, r % lcm);
                if (currentL <= currentR) ans = currentR - currentL + 1;
            } else {
                ans = ((beforeR - beforeL - 1) / lcm) * mx;
                ans += min(r % lcm + 1, mx);
                if (l % lcm < mx) {
                    ans += mx - (l % lcm) - 1;
                }

            }

            ll size = r - l + 1;
            cout << (size - ans) << " ";
        }
        cout << endl;
    }
}
