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;
class gfg {
 public:
  int gcd(int a, int b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  int lcm(int a, int b) { return (a * b) / gcd(a, b); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int a, b, q;
    cin >> a >> b >> q;
    gfg g;
    int n = g.lcm(a, b);
    vector<int> num(n, 0);
    int temp = min(a, b);
    int mod1, mod2;
    for (int i = temp; i < temp + n; i++) {
      mod1 = (i % a) % b;
      mod2 = (i % b) % a;
      if (mod1 != mod2) num[i - temp] = 1;
    }
    for (int i = 1; i < n; i++) num[i] += num[i - 1];
    for (int i = 0; i < q; i++) {
      long long l, r;
      cin >> l >> r;
      if (i != 0) cout << " ";
      if (l < (long long)temp && r < (long long)temp)
        cout << "0";
      else {
        if (l < (long long)temp && r >= (long long)temp) l = temp;
        long long total = (long long)(r + 1 - l) / n;
        long long ans = 0;
        ans += (long long)total * num[n - 1];
        l = (l - (long long)temp) % n;
        r = (r - (long long)temp) % n;
        if (r >= l) {
          ans += num[r];
          if (l != 0) ans -= num[l - 1];
        } else {
          ans += num[n - 1] - num[l - 1];
          if (r != 0) ans -= num[r - 1];
        }
        cout << ans;
      }
    }
    cout << endl;
  }
  return 0;
}
