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 lli long long int
#define fi first
#define se second
#define pb push_back
#define forn(i,n) for(int i = 0; i < n; i++)
#define forr(i,a,n) for(int i = a; i<n; i++)
#define fastIO(); ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mp make_pair
#define ld long double
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<lli> vlli;
int t, q, a, b;
lli l, r, ans, difMinComMul, minComMul;
void precal(vlli& val){
	forn(i,minComMul){
		//cout << (i%a)%b << ' ' << (i%b)%a << "  gg\n";
		if((i%a)%b == (i%b)%a){
			val[i] = 0;
		}
		else{
			difMinComMul++;
			val[i] = 1;
		}
	}
}

void solve(){
	cin >>a >>b>> q;
	ans = 0;
	minComMul = (a*b)/(__gcd(a,b));
	vlli val (minComMul,0);
	difMinComMul = 0;
	precal(val);
	//forn(i,minComMul) cout << val[i] << ' '; cout << '\n';
	if(a == b){
		forn(i,q) cin >> l >> r;
		cout << "0 ";
	}
	else{
		forn(i,q){
			ans = 0;
			cin >> l >> r;
			ans += ((r-l+1)/minComMul)*difMinComMul;
			int falta = (r-l+1)%minComMul;
			forn(i,falta){
				if(val[l%minComMul]) ans++;
				l++;
			}
			cout << ans << ' ';
		}
	}
	cout << '\n';
}
int main(){
	fastIO();
	cin >> t;
	while(t--) solve();
	return 0;
}
