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 all(x) (x).begin(), (x).end()
using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<vi> vvi;

const int mod = 998244353;
const int inf = 2e9 + 5;

int main() {
    ios::sync_with_stdio(false);

    int t;
    cin >> t;

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

        if (a > b) swap(a, b);
        ll g = __gcd(a, b);

        if (b == (a * b) / g) {

            for (int i = 0; i < q; i++){
                int l, r; cin >> l >> r;
                cout << "0 ";
            }
            cout << "\n";
            continue;
        }

        while (q--) {
            ll l, r;
            cin >> l >> r;

            int l1 = l, r1 = r;

            ll ans = r - l + 1;
            if (l < b and r > b)
                ans -= b - l;
            else if (r < b) {
                cout << "0 ";
                continue;
            }

            ll num = a * b / g;

            if (l % num != 0)
                l = l + num - l % num;
            r = r - r % num;

            l /= num;
            r /= num;

            ans -= (r - l) * b;
            if (r * num + b - 1 < r1)
                ans -= b;
            else ans -= r1 - r * num + 1;
            cout << ans << " ";
        }
        cout << "\n";
    }
    return 0;
}