Problem: You are given an array a consisting of n integers. You can perform the following operations with it: 

  1. Choose some positions i and j (1 ≤ i, j ≤ n, i ≠ j), write the value of a_i ⋅ a_j into the j-th cell and remove the number from the i-th cell; 
  2. Choose some position i and remove the number from the i-th cell (this operation can be performed no more than once and at any point of time, not necessarily in the beginning). 



The number of elements decreases by one after each operation. However, the indexing of positions stays the same. Deleted numbers can't be used in the later operations.

Your task is to perform exactly n - 1 operations with the array in such a way that the only number that remains in the array is maximum possible. This number can be rather large, so instead of printing it you need to print any sequence of operations which leads to this maximum number. Read the output format to understand what exactly you need to print.

Input

The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.

The second line contains n integers a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9) — the elements of the array.

Output

Print n - 1 lines. The k-th line should contain one of the two possible operations.

The operation of the first type should look like this: 1~ i_k~ j_k, where 1 is the type of operation, i_k and j_k are the positions of the chosen elements.

The operation of the second type should look like this: 2~ i_k, where 2 is the type of operation, i_k is the position of the chosen element. Note that there should be no more than one such operation.

If there are multiple possible sequences of operations leading to the maximum number — print any of them.

Examples

Input

5
5 -2 0 1 -3


Output

2 3
1 1 2
1 2 4
1 4 5


Input

5
5 2 0 4 0


Output

1 3 5
2 5
1 1 2
1 2 4


Input

2
2 -1


Output

2 2


Input

4
0 -10 0 0


Output

1 1 2
1 2 3
1 3 4


Input

4
0 0 0 0


Output

1 1 2
1 2 3
1 3 4

Note

Let X be the removed number in the array. Let's take a look at all the examples:

The first example has, for example, the following sequence of transformations of the array: [5, -2, 0, 1, -3] → [5, -2, X, 1, -3] → [X, -10, X, 1, -3] → [X, X, X, -10, -3] → [X, X, X, X, 30]. Thus, the maximum answer is 30. Note, that other sequences that lead to the answer 30 are also correct.

The second example has, for example, the following sequence of transformations of the array: [5, 2, 0, 4, 0] → [5, 2, X, 4, 0] → [5, 2, X, 4, X] → [X, 10, X, 4, X] → [X, X, X, 40, X]. The following answer is also allowed: 
    
    
      
    1 5 3  
    1 4 2  
    1 2 1  
    2 3  
    

Then the sequence of transformations of the array will look like this: [5, 2, 0, 4, 0] → [5, 2, 0, 4, X] → [5, 8, 0, X, X] → [40, X, 0, X, X] → [40, X, X, X, X].

The third example can have the following sequence of transformations of the array: [2, -1] → [2, X].

The fourth example can have the following sequence of transformations of the array: [0, -10, 0, 0] → [X, 0, 0, 0] → [X, X, 0, 0] → [X, X, X, 0].

The fifth example can have the following sequence of transformations of the array: [0, 0, 0, 0] → [X, 0, 0, 0] → [X, X, 0, 0] → [X, X, X, 0].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
const long long int INF = LONG_LONG_MAX;
const vector<vector<long long int> > adj4({{0, 1}, {0, -1}, {1, 0}, {-1, 0}});
const vector<vector<long long int> > adj8(
    {{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {-1, -1}, {1, -1}, {-1, 1}, {1, 1}});
struct DisjointSet {
  vector<long long int> roots;
  vector<long long int> sizes;
  DisjointSet(long long int n) {
    sizes = vector<long long int>(n, 1);
    roots = vector<long long int>(n);
    for (long long int i = 0; i < (n); ++i) roots[i] = i;
  }
  long long int root(long long int a) {
    return roots[a] == a ? a : root(roots[a]);
  }
  void join(long long int a, long long int b) {
    long long int ra = root(a);
    long long int rb = root(b);
    if (sizes[a] > sizes[b]) {
      roots[rb] = ra;
      sizes[ra] += sizes[rb];
    } else {
      roots[ra] = rb;
      sizes[rb] += sizes[ra];
    }
  }
};
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
bool isprime(long long int n) {
  if (n == 2 || n == 3) return true;
  if (n < 2 || n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 6; (i - 1) * (i - 1) <= n; i += 6) {
    if (n % (i - 1) == 0 || n % (i + 1) == 0) {
      return false;
    }
  }
  return true;
}
vector<bool> sprime;
void genPrime(long long int sz) {
  sprime = vector<bool>(sz, true);
  sprime[0] = false;
  sprime[1] = false;
  for (long long int i = (2); i < (sz); ++i) {
    for (long long int j = i * i; j < sz; j += i) {
      sprime[j] = false;
    }
  }
}
vector<long long int> facts;
bool factGenFlag = false;
void genFact(long long int n) {
  factGenFlag = true;
  facts = vector<long long int>(n + 1);
  facts[0] = 1;
  for (long long int i = (1); i < (n + 1); ++i) {
    facts[i] = facts[i - 1] * i;
  }
}
void genFactMod(long long int n, long long int mod) {
  factGenFlag = true;
  facts = vector<long long int>(n + 1);
  facts[0] = 1;
  for (long long int i = (1); i < (n + 1); ++i) {
    facts[i] = facts[i - 1] * i % mod;
  }
}
long long int fact(long long int a, long long int n) {
  if (factGenFlag && a < 3) return facts[n];
  long long int r = 1;
  for (long long int i = (a); i < (n + 1); ++i) {
    r *= i;
  }
  return r;
}
long long int factMod(long long int a, long long int n, long long int mod) {
  if (factGenFlag && a < 3) return facts[n];
  long long int r = 1;
  for (long long int i = (a); i < (n + 1); ++i) {
    r *= i;
    r %= mod;
  }
  return r;
}
long long int powMod(long long int a, long long int b, long long int mod) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= mod;
    }
    p *= p;
    p %= mod;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a, long long int mod) {
  return powMod(a, mod - 2, mod);
}
long long int ent(long long int a, long long int b) {
  long long int r = fact(a - b + 1, a);
  long long int d = fact(2, b);
  long long int ent = r / d;
  return ent;
}
long long int entMod(long long int a, long long int b, long long int mod) {
  long long int r = factMod(a - b + 1, a, mod);
  long long int d = factMod(2, b, mod);
  long long int ent = r * modularInverse(d, mod) % mod;
  return ent;
}
long long int bsfunction(long long int n) { return 0; }
long long int binarysearch(long long int left, long long int right,
                           long long int target) {
  if (left == right) {
    return left;
  }
  long long int mid =
      (left + right > 0) ? (left + right) / 2 : (left + right - 1) / 2;
  long long int val = bsfunction(mid);
  if (val > target) {
    return binarysearch(left, mid, target);
  } else if (val == target) {
    return mid;
  } else {
    return binarysearch(mid + 1, right, target);
  }
}
vector<long long int> topores;
vector<bool> topovisperm;
vector<bool> topovistemp;
vector<vector<long long int> > topograph;
bool topoimp;
void toposortvisit(long long int step) {
  if (topovisperm[step]) return;
  if (topovistemp[step]) {
    topoimp = true;
    return;
  }
  topovistemp[step] = true;
  for (long long int p : topograph[step]) {
    toposortvisit(p);
    if (topoimp) return;
  }
  topovisperm[step] = true;
  topores.push_back(step);
}
void toposort() {
  long long int n = topograph.size();
  topoimp = false;
  topovistemp = vector<bool>(n, false);
  topovisperm = vector<bool>(n, false);
  topores = vector<long long int>();
  topores.reserve(n);
  for (long long int i = 0; i < (n); ++i) {
    if (!topovisperm[i]) {
      toposortvisit(i);
      if (topoimp) {
        return;
      }
    }
  }
}
stringstream sss;
void MAIN() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  for (long long int i123458 = 0; i123458 < (n); ++i123458) cin >> arr[i123458];
  long long int x = 0;
  long long int y = -1;
  long long int lst = -1;
  vector<bool> xx(n);
  for (long long int i = 0; i < (n); ++i) {
    long long int a = arr[i];
    if (a < 0) {
      x = 1 - x;
      if (y == -1 || -a < -arr[y]) {
        y = i;
      }
    }
    if (a == 0) {
      lst = i;
    }
  }
  if (x == 1) {
    if (lst == -1) {
      cout << "2 " << y + 1 << endl;
    } else {
      cout << "1 " << y + 1 << ' ' << lst + 1 << endl;
    }
    xx[y] = true;
  }
  if (lst != -1) {
    long long int c = 0;
    for (long long int i = 0; i < (n); ++i) {
      if (c > n - 3 - x || i == lst) break;
      if (arr[i] == 0) {
        ++c;
        cout << "1 " << i + 1 << ' ' << lst + 1 << endl;
        xx[i] = true;
      }
    }
    if (!x || n > 2) cout << "2 " << lst + 1 << endl;
    xx[lst] = true;
  }
  long long int ls = n - 1;
  while (ls >= 0 && xx[ls]) --ls;
  if (ls >= 0) {
    for (long long int i = 0; i < (n); ++i) {
      if (i == ls) break;
      if (!xx[i]) {
        cout << "1 " << i + 1 << ' ' << ls + 1 << endl;
      }
    }
  }
}
const long long int TESTCASEN = 1;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  sss << R"(
5
5 2 0 1 -3
2
2 -1
4
0 -10 0 0
4
0 0 0 0
    )";
  for (long long int i = 0; i < (TESTCASEN); ++i) {
    MAIN();
  }
  return 0;
}