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>
using namespace std;
typedef vector< int > vi;
typedef vector< vi > vvi;
typedef pair< int, int > ii;
#define ll long long
#define MOD 1000000007
#define dd double
#define f0(i, n) for(int i = 0; i < n; i++)
#define fab(i,a,b) for(int i=a;i<b;i++)
#define f1(i,b) for(int i=1;i<=b;i++)
#define pb push_back
#define mp make_pair
#define clr(x) x.clear()
#define sz(x) ((int)(x).size())
#define F first
#define S second
#define all(c) c.begin(), c.end()
#define allr(c) c.rbegin(), c.rend()


ll func(ll x, ll lcm, ll b) {

	ll div = x / lcm;
	ll prod = div * lcm;
	ll ans = x;

	if (div != 0) {

		ans -= min(b, x - prod + 1);
		ans -= ((div - 1) * b);
	}
	return ans - min(x, (b - 1));


}


int main() {


	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif


//ios_base& scientific (ios_base& str);
	int t;
	cin >> t;

	while (t--) {

		int a, b, q;
		cin >> a >> b >> q;

		if (a > b) {
			swap(a, b);
		}

		ll lcm = (a * b) / (__gcd(a, b));
		// cout << lcm << endl;

		while (q--) {
			ll cnt;

			ll l, r;
			cin >> l >> r;

			if (l == r) {
				cout << 0 << " ";
			}
			else {

				if (lcm == b || r < b) {
					cout << 0 << " ";
					continue;
				}

				cnt = func(r, lcm, b);

				if (l > 1) {
					cnt -= func(l - 1, lcm, b);
				}
				cout << cnt << " ";

			}



		}
		cout << endl;
	}




	return 0;
}