Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long INF = 1e7;
double EPS = 1e-12;
long double tick() {
  static clock_t oldt;
  clock_t newt = clock();
  long double diff = 1.0L * (newt - oldt) / CLOCKS_PER_SEC;
  oldt = newt;
  return diff;
}
inline long long int sc() {
  long long int a;
  scanf("%lld", &a);
  return a;
}
inline int scd() {
  int a;
  scanf("%d", &a);
  return a;
}
inline double scf() {
  double a;
  scanf("%lf", &a);
  return a;
}
template <typename T1>
void printIt(T1 t1) {
  cout << t1 << " )" << endl;
}
template <typename T1, typename... T2>
void printIt(T1 t1, T2... t2) {
  cout << t1 << " , ";
  printIt(t2...);
}
long long int powP(long long int x, long long int y) {
  long long int ans = 1, po = x % mod;
  while (y > 0) {
    if (y & 1) ans = ans * po % mod;
    y >>= 1;
    po = po * po % mod;
  }
  return ans;
}
bool bitSet(long long int n, long long int bit) {
  if ((n & (1LL << bit)) != 0)
    return true;
  else
    return false;
}
long long int findGcd(long long int a, long long int b) {
  while (a != 0) {
    long long int temp = b % a;
    b = a;
    a = temp;
  }
  return b;
}
long long int find(long long int x, long long int a, long long int b) {
  long long int comm = a * b / findGcd(a, b);
  long long int times = x / comm, rem = x % comm;
  long long int val = times * b + min(rem, b - 1);
  return val;
}
int main() {
  long long int T = sc();
  while (T--) {
    long long int a = sc(), b = sc(), q = sc();
    if (a > b) swap(a, b);
    while (q--) {
      long long int l = sc(), r = sc();
      long long int eq = find(r, a, b) - find(l - 1, a, b);
      printf("%lld ", (r - l + 1) - eq);
    }
    printf("\n");
  }
  return 0;
}