Problem: Ujan has finally cleaned up his house and now wants to decorate the interior. He decided to place a beautiful carpet that would really tie the guest room together.

He is interested in carpets that are made up of polygonal patches such that each side of a patch is either a side of another (different) patch, or is an exterior side of the whole carpet. In other words, the carpet can be represented as a planar graph, where each patch corresponds to a face of the graph, each face is a simple polygon. The perimeter of the carpet is the number of the exterior sides. 

Ujan considers a carpet beautiful if it consists of f patches, where the i-th patch has exactly a_i sides, and the perimeter is the smallest possible. Find an example of such a carpet, so that Ujan can order it!

Input

The first line of input contains a single integer f (1 ≤ f ≤ 10^5), the number of patches in the carpet. 

The next line contains f integers a_1, …, a_f (3 ≤ a_i ≤ 3⋅ 10^5), the number of sides of the patches. The total number of the sides of the patches a_1 + … + a_f does not exceed 3⋅10^5.

Output

Output the description of the carpet as a graph. 

First, output a single integer n (3 ≤ n ≤ 3 ⋅ 10^5), the total number of vertices in your graph (the vertices must be numbered from 1 to n). 

Then output f lines containing the description of the faces. The i-th line should describe the i-th face and contain a_i distinct integers v_{i,1}, …, v_{i,a_i} (1 ≤ v_{i,j} ≤ n), which means that the vertices v_{i,j} and v_{i,(j mod{a_i})+1} are connected by an edge for any 1 ≤ j ≤ a_i.

The graph should be planar and satisfy the restrictions described in the problem statement. Its perimeter should be the smallest possible. There should be no double edges or self-loops in the graph. The graph should be connected. Note that a solution always exists; if there are multiple solutions, output any of them.

Examples

Input


2
3 3


Output


4
2 1 4 
1 2 3 


Input


3
5 3 5


Output


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

Note

In the first sample, the two triangular faces are connected by a single edge, which results in the minimum perimeter 4.

The figure shows one possible configuration for the second sample. The minimum perimeter in this case is 3. 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int f;
  cin >> f;
  vector<pair<int, int>> a(f);
  for (int i = 0; i < f; ++i) {
    int x;
    cin >> x;
    a[i] = {x, i};
  }
  sort(a.begin(), a.end(), greater<pair<int, int>>());
  int n = 0;
  deque<int> q;
  vector<vector<int>> v(f);
  for (int i = 0; i < a[0].first; ++i) {
    q.push_back(n);
    v[a[0].second].push_back(n);
    ++n;
  }
  for (int i = 1; i < f; ++i) {
    int e, k;
    tie(e, k) = a[i];
    int p =
        min(e - 1, ((int)q.size() + e - (i == f - 1 ? 3 : a[i + 1].first)) / 2);
    for (int j = 0; j < p - 1; ++j) {
      v[k].push_back(q.back());
      q.pop_back();
    }
    v[k].push_back(q.back());
    for (int j = p; j < e - 1; ++j) {
      v[k].push_back(n);
      q.push_back(n);
      ++n;
    }
    v[k].push_back(q.front());
    q.push_back(q.front());
    q.pop_front();
  }
  cout << n << "\n";
  for (int i = 0; i < f; ++i)
    for (int j : v[i]) cout << j + 1 << " \n"[j == v[i].back()];
  return 0;
}