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 double PI = acos(-1.0);
int dx4[] = {+1, -1, +0, +0};
int dy4[] = {+0, +0, +1, -1};
int dx8[] = {+0, +0, -1, +1, -1, -1, +1, +1};
int dy8[] = {-1, +1, +0, +0, -1, +1, -1, +1};
int dr[] = {-1, 1, -2, -2, -1, 1, 2, 2};
int dc[] = {-2, -2, -1, 1, 2, 2, 1, -1};
bool isNotEqual(int x, int a, int b) { return (x % a) % b != (x % b) % a; }
long long calculate(long long val, long long tot, long long maxi,
                    long long cnt) {
  long long div = val / tot;
  long long rem = val % tot;
  long long ans = div * cnt;
  rem = rem - (maxi - 1);
  if (rem > 0) ans += rem;
  return ans;
}
void test_case() {
  int a, b;
  cin >> a >> b;
  int maxi = max(a, b);
  int mini = min(a, b);
  int cnt = 0;
  for (int i = maxi; isNotEqual(i, a, b); ++i) cnt++;
  int tot = maxi + cnt;
  int query;
  cin >> query;
  while (query-- > 0) {
    long long lft, rgt;
    cin >> lft >> rgt;
    cout << calculate(rgt, tot, maxi, cnt) - calculate(lft - 1, tot, maxi, cnt)
         << " \n"[query == 0];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(15);
  cout << fixed;
  int T = 1;
  cin >> T;
  for (int tc = 1; tc <= T; ++tc) {
    test_case();
  }
  return 0;
}
