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 faster_io ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define mem(a,b) memset(a, b, sizeof a)
#define F first
#define S second
#define ll long long
#define pb push_back
#define pll pair<ll, ll>
using namespace std;
const ll N = 1e6 + 1e5, inf = 3e18, mod = 1e9 + 7;
ll tc, n, m, q, a, b, c, d, i, j, k, l, r, L, R;
ll lft[N], rgt[N];
vector <ll> vec;
string t;

int main()
{
    faster_io;
    cin >> tc;
    while(tc--)
    {
        cin >> a >> b >> q;
        c = a * b / __gcd(a, b);
        for(i = 0; i <= c + 1; i++)
            lft[i] = rgt[i] = 0;
        for(i = 1; i <= c; i++)
            lft[i] = lft[i - 1] + (((i % a) % b) != ((i % b) % a));
        for(i = c; i > 0; i--)
            rgt[i] = rgt[i + 1] + (((i % a) % b) != ((i % b) % a));
        while(q--)
        {
            cin >> L >> R;
            i = R - L + 1;
            j = i / c;
            ll ans = lft[c] * j;
            if(i - j * c + L % c <= c)
                cout << ans + lft[i - j * c + L % c - 1] - lft[L % c - 1] << ' ';
            else
                cout << ans + lft[c] - lft[L % c + 1] + rgt[i - c + L % c - 1] << ' ';
        }
        cout << endl;
    }
}
