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 long long mod = 1000000007, N = 1e5 + 5, M = 1e5 + 5, INF = 0x3f3f3f3f;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int main(void) {
  int T;
  scanf("%d", &T);
  while (T--) {
    int a, b, q;
    long long l, r;
    vector<long long> ans;
    scanf("%d%d%d", &a, &b, &q);
    int lcm = a * b / gcd(a, b);
    vector<int> sum(lcm + 1, 0), cnt(lcm + 1, 0);
    for (int i = 1; i < lcm + 1; i++) {
      if (i % a % b != i % b % a) {
        sum[i] = sum[i - 1] + 1;
        cnt[i] = 1;
      } else
        sum[i] = sum[i - 1];
    }
    for (int i = 0; i < q; i++) {
      long long now = 0;
      scanf("%lld%lld", &l, &r);
      if (l % lcm) {
        long long up = (l / lcm * lcm + lcm);
        if (up >= r) {
          now += sum[r % lcm] - sum[(l % lcm) - 1];
          l = r + 1;
        } else {
          now += sum[lcm] - sum[l % lcm - 1];
          l = up;
        }
      }
      if (r % lcm && r >= l) {
        long long down = (r / lcm * lcm);
        now += sum[r % lcm];
        r = down;
      }
      now += sum[lcm] * ((r - l) / lcm);
      ans.emplace_back(now);
    }
    for (int i = 0; i < q; i++) printf("%lld%c", ans[i], " \n"[i == q - 1]);
  }
  return 0;
}
