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 T>
inline T bigMod(T p, T e, T M) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T modInverse(T a, T M) {
  return bigMod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T>
inline T SQR(T a) {
  return a * a;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long int a, b, q;
    scanf("%lld %lld", &a, &b);
    scanf("%lld", &q);
    ;
    if (a > b) swap(a, b);
    long long int lc = lcm(a, b);
    while (q--) {
      long long int x, y;
      scanf("%lld %lld", &x, &y);
      if (y < b)
        printf("0 ");
      else if (y == b) {
        if (lc == b)
          printf("0 ");
        else
          printf("1 ");
      } else {
        long long int total = 0;
        long long int c = (x / lc);
        long long int d = (y / lc);
        if (c == d) {
          if (c * lc + b >= x) total += min(y, c * lc + b) - x + 1;
        } else {
          total += (d - c - 1) * b;
          if (d * lc + b > y)
            total += (y - d * lc + 1);
          else
            total += b;
          if (c * lc + b >= x) total += (c * lc + b + 1 - x);
        }
        if (!c) total--;
        total = y - x + 1 - total;
        printf("%lld", total);
        printf(" ");
      }
    }
    printf("\n");
  }
  return 0;
}
