Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (long long)((1e5) + 100);
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void pr(long long x) { cout << x; }
void prl(long long x) { cout << x << "\n"; }
long long cbrt(long long x) {
  long long lo = 1, hi = min(2000000ll, x);
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    if (mid * mid * mid < x) {
      lo = mid;
    } else
      hi = mid;
  }
  if (hi * hi * hi <= x)
    return hi;
  else
    return lo;
}
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
long long XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const long long N = (long long)(2 * 1e5 + 10);
long long findcount(long long r, long long lcm, long long b) {
  long long div = r / lcm;
  long long prod = div * lcm;
  long long tot = r;
  if (div != 0) {
    tot -= min(b, r - prod + 1);
    tot -= (div - 1) * b;
  }
  return tot - min(r, (b - 1));
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) {
    long long a, b, q;
    cin >> a >> b >> q;
    if (a > b) swap(a, b);
    long long lc = a * b;
    long long gg = gcd(a, b);
    lc /= gg;
    while (q--) {
      long long l, r;
      cin >> l >> r;
      if (lc == b || (r < b)) {
        cout << 0 << " ";
        continue;
      }
      long long cnt = findcount(r, lc, b);
      if (l > 1) cnt -= findcount(l - 1, lc, b);
      cout << cnt << " ";
    }
    cout << "\n";
  }
}