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;
long long t, a, b, q, l, r, l0, r0, mi, ma, tot, ans, x[40005];
long long calc2(long long r) {
  int pos = lower_bound(x + 1, x + tot + 1, r) - x;
  return x[pos] == r ? pos : pos - 1;
}
long long calc1(long long l) {
  return tot + 1 - (lower_bound(x + 1, x + tot + 1, l) - x);
}
int main() {
  for (scanf("%lld", &t); t--;) {
    scanf("%lld%lld%lld", &a, &b, &q);
    mi = a * b + 1, ma = -1, tot = 0;
    for (long long i = 1; i <= a * b; i++)
      if (i % a % b != i % b % a)
        mi = min(mi, i), ma = max(ma, i), x[++tot] = i;
    while (q--) {
      scanf("%lld%lld", &l, &r);
      if (ma == -1) {
        printf("0 ");
        continue;
      }
      l0 = l / (a * b), r0 = r / (a * b), l %= (a * b), r %= (a * b);
      printf("%lld ", (r0 - l0 - 1) * tot + calc1(l) + calc2(r));
    }
    puts("");
  }
  return 0;
}
