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 MAXN = 1e5 + 5;
const long long MOD = 1e9 + 7;
int N, M, K, Q;
int a, b, q;
long long psum[MAXN];
long long query(long long n) {
  long long m = (long long)a * (long long)b;
  return n / m * psum[m] + psum[n % m];
}
void solve(int testNum) {
  int i, j;
  scanf("%d%d%d", &a, &b, &q);
  for (int i = 1; i <= a * b; ++i) {
    int x = i;
    psum[x] = psum[x - 1];
    if (x % a % b != x % b % a) ++psum[x];
  }
  long long l, r;
  for (i = 1; i <= q; ++i) {
    scanf("%lld%lld", &l, &r);
    long long ans = query(r) - query(l - 1LL);
    printf("%lld ", ans);
  }
  printf("\n");
  return;
}
int main() {
  int T;
  scanf("%d", &T);
  for (int t = 1; t <= T; ++t) {
    solve(t);
  }
  return 0;
}
