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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
struct _ {
  ios_base::Init i;
  _() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); }
} _;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> m(n + 1, INT_MAX);
  for (int i = 1; i <= n; i++) cin >> m[i];
  vector<int> c(k + 1, 0);
  for (int i = 1; i <= k; i++) cin >> c[i];
  ;
  sort(m.begin(), m.end(), greater<int>());
  int min_diff = 0, curr_diff = 0;
  vector<int> ans;
  vector<vector<int>> v;
  for (int i = 1; i <= n; i++) {
    if (min_diff == 0) {
      v.push_back(ans);
      ans.clear();
      min_diff = c[m[i]];
      curr_diff = c[m[i]];
    }
    if (m[i] != m[i - 1]) curr_diff = c[m[i]];
    ;
    ans.push_back(m[i]);
    min_diff--;
    curr_diff--;
    min_diff = min(min_diff, curr_diff);
  }
  v.push_back(ans);
  cout << v.size() - 1 << '\n';
  for (int i = 1; i <= v.size() - 1; i++) {
    cout << v[i].size() << " ";
    for (auto j : v[i]) cout << j << " ";
    cout << '\n';
  }
  return 0;
}
