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 <cmath>
#include <functional>
#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <chrono>
#include <cstring>
#include <cstdlib>

using namespace std;
 
typedef long long ll;

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while(t--) {
        long long a, b, q;
        cin >> a >> b >> q;
        if(a < b) swap(a, b);
        long long lcm = a*b/gcd(a, b);
        for(int lo = 0; lo < q; lo++) {
            long long l, r, ans = 0;
            cin >> l >> r;
            long long last = r-(r%lcm), first = l%lcm == 0? l : l - (l%lcm) + lcm;
            for(long long i = l; i < a && i <= r; i++) ans++;
            if(last > first)
                ans += (last-first)*a/lcm;
            if(last >= l) 
                for(long long i = last; i < last+a && i <= r; i++) ans++;
            cout << r-l+1 - ans << " \n"[lo+1==q];
        }
    }
    return 0;
}