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;
unsigned long long mod = 1e9 + 7;
unsigned long long ncr(long long int n, long long int k) {
  unsigned long long res = 1;
  if (k > n - k) k = n - k;
  for (int i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
    res = res;
  }
  return res;
}
bool prime(unsigned long long num) {
  bool flag = true;
  for (unsigned long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      flag = false;
      break;
    }
  }
  return flag;
}
long long ipow(long long base, long long exp) {
  long long result = 1;
  for (;;) {
    if (exp & 1) {
      result *= base;
      result = result % mod;
    }
    exp >>= 1;
    if (!exp) break;
    base *= base;
    base = base % mod;
  }
  return result % mod;
}
long long abs(long long a, long long b) {
  if ((a - b) > 0)
    return (a - b);
  else
    return (b - a);
}
long long MOD = 998244353;
long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long der(int k) {
  long long res = 0;
  long long fact = 1;
  for (int i = 2; i <= k; i++) fact *= i;
  int s = 1;
  long long j = 1;
  for (int i = 2; i <= k; i++) {
    j = j * i;
    long long first = (fact / j) * s;
    res += first;
    s = s * -1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long h = (a * b) / gcd(a, b);
    long long ind = h / b;
    long long eq = b;
    long long neq = h - eq;
    if (h == b) {
      neq = 0;
      eq = h;
    }
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (neq == 0) {
        cout << "0" << endl;
        continue;
      }
      long long left = (l - 1) / h;
      long long lefteq = min((l - 1) % h, b - 1);
      long long leftneq = (l - 1) % h - lefteq;
      lefteq += left * eq;
      leftneq += left * neq;
      long long rht = r / h;
      long long req = min(r % h, b - 1);
      long long rneq = r % h - req;
      req += rht * eq;
      rneq += rht * neq;
      cout << rneq - leftneq << " ";
    }
    cout << endl;
  }
}