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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int &n, int b) { n |= two(b); }
inline void unset_bit(int &n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
const double pi = acos(-1);
int cmp(double x, double y = 0, double tol = 1e-9) {
  return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
int main() {
  int TC;
  scanf("%d", &TC);
  while (TC--) {
    int a, b, q;
    scanf("%d%d%d", &a, &b, &q);
    vector<int> V(a * b + 1);
    for (int num = 1; num <= int(a * b); num++) {
      if ((num % a) % b != (num % b) % a) V[num]++;
      V[num] += V[num - 1];
    }
    while (q--) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      l--;
      int ab = a * b;
      long long ans = (r / ab) * V[ab] + V[r % ab];
      ans -= ((l / ab) * V[ab] + V[l % ab]);
      printf("%lld ", ans);
    }
    printf("\n");
  }
  return 0;
}
