#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
const int PRECISION = 20;
const int MOD = 1e9 + 7;
struct node {
  long long val;
  vector<long long> formula;
  node() { val = -1; }
};
struct group {
  long long mul, last, gcd;
  group(long long m, long long l, long long lm) {
    mul = m;
    last = l;
    gcd = lm;
  }
};
bool comp(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
int bs(vector<int> vc, int target) {
  int low = 0, high = vc.size() - 1;
  while (high > low) {
    int mid = ((high - low + 1) / 2) + low;
    if (vc[mid] <= target)
      low = mid;
    else
      high = mid - 1;
  }
  return low;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long fastPow(long long a, long long n, long long m) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long x = fastPow(a, n / 2, m) % m;
  x = (x * x) % m;
  return (n % 2 == 0 ? x : (x * a) % m);
}
int char2index(char a) { return a >= 'a' ? a - 'a' : a - 'A' + 26; }
int flipCase(char a) {
  return a >= 'a' ? char2index(a - ('a' - 'A')) : char2index(a + ('a' - 'A'));
}
void prefix(vector<long long> &arr) {
  for (int i = 1; i < arr.size(); i++) arr[i] += arr[i - 1];
}
long long eval(vector<long long> &coeffs, long long x) {
  long long ans = 0;
  long long xs = 1;
  for (long long i = coeffs.size() - 1; i >= 0; i--) {
    ans += (coeffs[i] * xs);
    xs *= x;
  }
  return ans;
}
string solve(int n, int k, string s) {
  vector<long long> pref(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + (s[i - 1] == '1' ? 1 : 0);
  }
  long long deno = (long long)n * n;
  long long nemo = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '0') continue;
    long long ans = (pref[i] - (i > k ? pref[i - k - 1] : 0)) * 2 - 1;
    nemo += ans;
  }
  long long g = gcd(deno, nemo);
  return to_string((nemo / g)) + "/" + to_string((deno / g));
}
int bracketToInt(char a) {
  switch (a) {
    case '(':
      return 1;
    case ')':
      return -1;
    case '{':
      return 2;
    case '}':
      return -2;
    case '[':
      return 3;
    case ']':
      return -3;
    case '<':
      return 4;
    case '>':
      return -4;
    default:
      return -1;
  }
}
void solve(long long n, long long p) {
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      if (ind == (2 * n) + p) break;
      cout << i + 1 << ' ' << j + 1 << '\n';
      ind++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, ee;
  ee = 0;
  cin >> t;
  while (t--) {
    long long n, p;
    cin >> n >> p;
    solve(n, p);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

Let's call an undirected graph of n vertices p-interesting, if the following conditions fulfill: 

  * the graph contains exactly 2n + p edges; 
  * the graph doesn't contain self-loops and multiple edges; 
  * for any integer k (1 ≤ k ≤ n), any subgraph consisting of k vertices contains at most 2k + p edges. 



A subgraph of a graph is some set of the graph vertices and some set of the graph edges. At that, the set of edges must meet the condition: both ends of each edge from the set must belong to the chosen set of vertices. 

Your task is to find a p-interesting graph consisting of n vertices.

Input

The first line contains a single integer t (1 ≤ t ≤ 5) — the number of tests in the input. Next t lines each contains two space-separated integers: n, p (5 ≤ n ≤ 24; p ≥ 0; <image>) — the number of vertices in the graph and the interest value for the appropriate test. 

It is guaranteed that the required graph exists.

Output

For each of the t tests print 2n + p lines containing the description of the edges of a p-interesting graph: the i-th line must contain two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — two vertices, connected by an edge in the resulting graph. Consider the graph vertices numbered with integers from 1 to n. 

Print the answers to the tests in the order the tests occur in the input. If there are multiple solutions, you can print any of them.

Examples

Input

1
6 0


Output

1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
const int PRECISION = 20;
const int MOD = 1e9 + 7;
struct node {
  long long val;
  vector<long long> formula;
  node() { val = -1; }
};
struct group {
  long long mul, last, gcd;
  group(long long m, long long l, long long lm) {
    mul = m;
    last = l;
    gcd = lm;
  }
};
bool comp(pair<int, int> p1, pair<int, int> p2) {
  return p1.second < p2.second;
}
int bs(vector<int> vc, int target) {
  int low = 0, high = vc.size() - 1;
  while (high > low) {
    int mid = ((high - low + 1) / 2) + low;
    if (vc[mid] <= target)
      low = mid;
    else
      high = mid - 1;
  }
  return low;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long fastPow(long long a, long long n, long long m) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long x = fastPow(a, n / 2, m) % m;
  x = (x * x) % m;
  return (n % 2 == 0 ? x : (x * a) % m);
}
int char2index(char a) { return a >= 'a' ? a - 'a' : a - 'A' + 26; }
int flipCase(char a) {
  return a >= 'a' ? char2index(a - ('a' - 'A')) : char2index(a + ('a' - 'A'));
}
void prefix(vector<long long> &arr) {
  for (int i = 1; i < arr.size(); i++) arr[i] += arr[i - 1];
}
long long eval(vector<long long> &coeffs, long long x) {
  long long ans = 0;
  long long xs = 1;
  for (long long i = coeffs.size() - 1; i >= 0; i--) {
    ans += (coeffs[i] * xs);
    xs *= x;
  }
  return ans;
}
string solve(int n, int k, string s) {
  vector<long long> pref(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1] + (s[i - 1] == '1' ? 1 : 0);
  }
  long long deno = (long long)n * n;
  long long nemo = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '0') continue;
    long long ans = (pref[i] - (i > k ? pref[i - k - 1] : 0)) * 2 - 1;
    nemo += ans;
  }
  long long g = gcd(deno, nemo);
  return to_string((nemo / g)) + "/" + to_string((deno / g));
}
int bracketToInt(char a) {
  switch (a) {
    case '(':
      return 1;
    case ')':
      return -1;
    case '{':
      return 2;
    case '}':
      return -2;
    case '[':
      return 3;
    case ']':
      return -3;
    case '<':
      return 4;
    case '>':
      return -4;
    default:
      return -1;
  }
}
void solve(long long n, long long p) {
  long long ind = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      if (ind == (2 * n) + p) break;
      cout << i + 1 << ' ' << j + 1 << '\n';
      ind++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t, ee;
  ee = 0;
  cin >> t;
  while (t--) {
    long long n, p;
    cin >> n >> p;
    solve(n, p);
  }
  return 0;
}

```