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>
constexpr int MOD = 1000000007;
using ll = long long;
using Pii = std::pair<int, int>;
using Pll = std::pair<ll, ll>;
namespace std {
template <class T>
string to_string(T s);
template <class S, class T>
string to_string(pair<S, T> p);
string to_string(string s) { return s; }
string to_string(const char s[]) { return to_string(string(s)); }
template <class T>
string to_string(T v) {
  if (v.empty()) return "{}";
  string ret = "{";
  for (auto x : v) ret += to_string(x) + ",";
  ret.back() = '}';
  return ret;
}
template <class S, class T>
string to_string(pair<S, T> p) {
  return "{" + to_string(p.first) + ":" + to_string(p.second) + "}";
}
}  // namespace std
void debug() { std::cerr << std::endl; }
template <class Head, class... Tail>
void debug(Head head, Tail... tail) {
  std::cerr << std::to_string(head) << " ";
  debug(tail...);
}
struct IO {
  std::string separator = " ";
  template <class T>
  inline void read(T& x) {
    char c;
    do {
      c = getchar_unlocked();
    } while (c != '-' && (c < '0' || '9' < c));
    bool minus = 0;
    if (c == '-') {
      minus = 1;
      c = getchar_unlocked();
    }
    x = 0;
    while ('0' <= c && c <= '9') {
      x *= 10;
      x += c - '0';
      c = getchar_unlocked();
    }
    if (minus) x = -x;
  }
  inline void read(std::string& x) {
    char c;
    do {
      c = getchar_unlocked();
    } while (c == ' ' || c == '\n');
    x.clear();
    do {
      x += c;
      c = getchar_unlocked();
    } while (c != ' ' && c != '\n' && c != EOF);
  }
  template <class T>
  inline void read(std::vector<T>& v) {
    for (auto& x : v) read(x);
  }
  template <class Head, class... Tail>
  inline void read(Head& head, Tail&... tail) {
    read(head);
    read(tail...);
  }
  template <class T>
  inline void write(T x) {
    char buf[32];
    int p = 0;
    if (x < 0) {
      x = -x;
      putchar_unlocked('-');
    }
    if (x == 0) putchar_unlocked('0');
    while (x > 0) {
      buf[p++] = (x % 10) + '0';
      x /= 10;
    }
    while (p) {
      putchar_unlocked(buf[--p]);
    }
  }
  inline void write(std::string x) {
    for (char c : x) putchar_unlocked(c);
  }
  inline void write(const char s[]) {
    for (int i = 0; s[i] != 0; ++i) putchar_unlocked(s[i]);
  }
  template <class T>
  inline void write(std::vector<T> v) {
    for (auto itr = v.begin(); itr + 1 != v.end(); ++itr) {
      write(*itr);
      write(separator);
    }
    write(v.back());
  }
  template <class Head, class... Tail>
  inline void write(Head head, Tail... tail) {
    write(head);
    write(separator);
    write(tail...);
  }
  template <class Head, class... Tail>
  inline void writeln(Head head, Tail... tail) {
    write(head, tail...);
    write("\n");
  }
  void set_separator(std::string s) { separator = s; }
} io;
using namespace std;
int main() {
  int t;
  io.read(t);
  while (t--) {
    int a, b, q;
    io.read(a, b, q);
    vector<int> table(a * b + 1);
    for (int i = 0; i < (int)a * b; i++) {
      table[i + 1] = table[i];
      if ((i % a) % b != (i % b) % a) table[i + 1]++;
    }
    ll sum = table[a * b];
    while (q--) {
      ll l, r;
      io.read(l, r);
      l--;
      ll ansl = sum * (l / (a * b)) + table[l % (a * b) + 1];
      ll ansr = sum * (r / (a * b)) + table[r % (a * b) + 1];
      io.write(ansr - ansl);
      io.write(q == 0 ? "\n" : " ");
    }
  }
  return 0;
}