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>
const int MAX_LEN = 1010;
using namespace std;
template <typename U, typename V>
string to_string(pair<U, V>);
string to_string(const string& e_) { return "\"" + e_ + "\""; }
string to_string(char e_) { return "\'" + string(1, e_) + "\'"; }
string to_string(bool e_) { return e_ ? "true" : "false"; }
template <typename T>
string to_string(T e_) {
  string s_ = "[ ";
  for (const auto& x_ : e_) s_ += to_string(x_) + " ";
  return s_ + "]";
}
template <typename U, typename V>
string to_string(pair<U, V> e_) {
  return "(" + to_string(e_.first) + ", " + to_string(e_.second) + ")";
}
void dbg_str() { ; }
template <typename U, typename... V>
void dbg_str(U u, V... v) {
  ;
  dbg_str(v...);
}
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * (b / gcd(a, b));
}
int t, q;
long long int a, b, l, r;
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld %lld", &a, &b);
    scanf("%d", &q);
    vector<long long int> till;
    long long int lc = lcm(a, b);
    till.assign(lc + 1, 0);
    till[0] = 0;
    for (int i = 1; i <= lc; i++) {
      till[i] = till[i - 1];
      if ((i % a) % b != (i % b) % a) till[i]++;
    }
    for (int i = 1; i <= lc; i++)
      ;
    while (q--) {
      long long int ans = 0;
      scanf("%lld %lld", &l, &r);
      l--;
      long long int left = (l / lc) * till[lc] + till[l % lc];
      long long int right = (r / lc) * till[lc] + till[r % lc];
      printf("%lld ", right - left);
    }
    printf("\n");
  }
}