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 long long ll;
typedef long double ld;
typedef pair < int, int > pii;
typedef pair < ll, ll > pll;

const int INF = 1000000007;
const ll INFLL = 1000000000000000007;

#define st first
#define nd second
#define pb push_back
#define _boost ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

// ***************************** CODE ***************************** //

int main() {
    _boost;

    int t; cin >> t;
    while(t--)
    {
        ll a, b, q;
        cin >> a >> b >> q;
        while(q--)
        {
            ll l, r;
            cin >> l >> r;

            ll nww = (a*b)/__gcd(a, b);
            ll i = 1, licz = 0;
            while(((i%a)%b) == ((i%b)%a))
            {
                i++;
                licz++;
                if(licz >= nww)
                    break;
            }
            ll ilea = l/nww;
            ll ileb = r/nww;
            ll ans = r-l+1;
            ans -= (ileb-ilea)*licz+1;
            if(ilea*nww+licz > l)
            {
                ans -= min(ilea*nww+licz-l, r-l+1);
            }
            if(ileb*nww+licz > r)
            {
                ans -= ileb*nww+licz-r;
            }
            cout << max((ll)0, ans) << " ";
        }
        cout << "\n";
    }

    return 0;
}
