#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1LL << 62;
const long long mx = 100005;
const double eps = 1e-10;
long long dx[10] = {1, 0, -1, 0}, dy[10] = {0, -1, 0, 1};
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long x = power(a, b / 2);
  x = x * x;
  if (b % 2) x = x * a;
  return x;
}
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2) % mod;
  x = (x * x) % mod;
  if (b % 2) x = (x * a) % mod;
  return x;
}
long long Set(long long N, long long pos) { return N = N | (1LL << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1LL << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1LL << pos)); }
int main() {
  long long tst, a, b, c, k, n, m, res = 0, ans = 0, t = 0;
  scanf("%lld", &tst);
  while (tst--) {
    scanf("%lld", &n);
    priority_queue<pair<long long, long long> > p;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a);
      p.push({a, i});
    }
    vector<pair<long long, long long> > v;
    while ((long long)(p).size() > 1) {
      pair<long long, long long> A = p.top();
      p.pop();
      pair<long long, long long> B = p.top();
      p.pop();
      if (A.first > 0 and B.first > 0) {
        v.push_back({A.second, B.second});
        A.first--;
        B.first--;
      }
      if (A.first) p.push(A);
      if (B.first) p.push(B);
    }
    cout << (long long)(v).size() << "\n";
    for (auto i : v) cout << i.first << " " << i.second << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

An important meeting is to be held and there are exactly n people invited. At any moment, any two people can step back and talk in private. The same two people can talk several (as many as they want) times per meeting.

Each person has limited sociability. The sociability of the i-th person is a non-negative integer a_i. This means that after exactly a_i talks this person leaves the meeting (and does not talk to anyone else anymore). If a_i = 0, the i-th person leaves the meeting immediately after it starts.

A meeting is considered most productive if the maximum possible number of talks took place during it.

You are given an array of sociability a, determine which people should talk to each other so that the total number of talks is as large as possible.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases.

The first line of each test case description contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) —the number of people in the meeting. The second line consists of n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2 ⋅ 10^5) — the sociability parameters of all people. 

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5. It is also guaranteed that the sum of all a_i (over all test cases and all i) does not exceed 2 ⋅ 10^5.

Output

Print t answers to all test cases.

On the first line of each answer print the number k — the maximum number of talks possible in a meeting.

On each of the next k lines print two integers i and j (1 ≤ i, j ≤ n and i ≠ j) — the numbers of people who will have another talk.

If there are several possible answers, you may print any of them.

Example

Input


8
2
2 3
3
1 2 3
4
1 2 3 4
3
0 0 2
2
6 2
3
0 0 2
5
8 2 0 1 1
5
0 1 0 0 6


Output


2
1 2
1 2
3
1 3
2 3
2 3
5
1 3
2 4
2 4
3 4
3 4
0
2
1 2
1 2
0
4
1 2
1 5
1 4
1 2
1
5 2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1LL << 62;
const long long mx = 100005;
const double eps = 1e-10;
long long dx[10] = {1, 0, -1, 0}, dy[10] = {0, -1, 0, 1};
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long x = power(a, b / 2);
  x = x * x;
  if (b % 2) x = x * a;
  return x;
}
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2) % mod;
  x = (x * x) % mod;
  if (b % 2) x = (x * a) % mod;
  return x;
}
long long Set(long long N, long long pos) { return N = N | (1LL << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1LL << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1LL << pos)); }
int main() {
  long long tst, a, b, c, k, n, m, res = 0, ans = 0, t = 0;
  scanf("%lld", &tst);
  while (tst--) {
    scanf("%lld", &n);
    priority_queue<pair<long long, long long> > p;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &a);
      p.push({a, i});
    }
    vector<pair<long long, long long> > v;
    while ((long long)(p).size() > 1) {
      pair<long long, long long> A = p.top();
      p.pop();
      pair<long long, long long> B = p.top();
      p.pop();
      if (A.first > 0 and B.first > 0) {
        v.push_back({A.second, B.second});
        A.first--;
        B.first--;
      }
      if (A.first) p.push(A);
      if (B.first) p.push(B);
    }
    cout << (long long)(v).size() << "\n";
    for (auto i : v) cout << i.first << " " << i.second << "\n";
  }
  return 0;
}

```