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;
const int INF = 0x3f3f3f3f;
const long long LINF = 4557430888798830399;
int a, b, q;
int c[80005];
void TestCase() {
  scanf("%d%d%d", &a, &b, &q);
  int T = a * b;
  for (int i = 1; i <= 2 * T; ++i) {
    int c1 = i % a % b;
    int c2 = i % b % a;
    c[i] = c[i - 1] + (c1 != c2);
  }
  while (q--) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    long long t = (r - l) / T;
    long long ans = t * c[T];
    l %= T;
    r %= T;
    if (r < l) r += T;
    ans += c[r] - (l >= 1 ? c[l - 1] : 0);
    printf("%lld ", ans);
  }
  puts("");
  return;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int tid = 1; tid <= t; ++tid) {
    TestCase();
  }
  return 0;
}
