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;
long long int q, a, b;
long long int mcm, con;
long long int sub[40040];
long long int gcd_recursive(long long int a, long long int b) {
  if (b)
    return gcd_recursive(b, a % b);
  else
    return a;
}
long long int check(long long int val) {
  if (((val % a) % b) != ((val % b) % a)) return 1;
  return 0;
}
long long int solve(long long int ini, long long int fin) {
  long long int sum = 0;
  if (fin - ini + 1 <= mcm) {
    if (fin % mcm >= ini % mcm)
      sum += sub[ini % mcm] - sub[(fin % mcm) + 1];
    else {
      sum += sub[ini % mcm];
      sum += sub[(fin % mcm)];
    }
  } else {
    if (ini % mcm != 1) {
      sum += sub[ini % mcm];
      long long int res = (ini % mcm);
      ini -= res;
      ini += mcm + 1;
    }
    long long int ran = fin - ini + 1;
    long long int div = ran / mcm;
    sum += con * div;
    ini += div * mcm;
    if (fin % mcm >= ini % mcm)
      sum += sub[ini % mcm] - sub[(fin % mcm) + 1];
    else {
      sum += sub[ini % mcm];
      sum += sub[(fin % mcm)];
    }
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  long long int t1, t2;
  while (t--) {
    cin >> a >> b >> q;
    mcm = a * b;
    mcm /= gcd_recursive(a, b);
    con = 0;
    memset(sub, 0, sizeof(sub));
    sub[mcm] = 0;
    for (int i = mcm - 1; i >= 0; i--) {
      con += check(i);
      sub[i] = check(i) + sub[i + 1];
    }
    while (q--) {
      cin >> t1 >> t2;
      cout << solve(t1, t2) << " ";
    }
    cout << endl;
  }
}
