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 n, cnt, kov;
vector<pair<int, int> > sz;
vector<int> sol[100002], p;
void pb() { p.pop_back(); }
void add(int a, int b) { sol[a].push_back(b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    sz.push_back({x, i});
  }
  sort(sz.rbegin(), sz.rend());
  for (int i = 1; i <= sz[0].first; i++)
    cnt++, p.push_back(i), add(sz[0].second, i);
  for (int i = 1; i < n; i++) {
    if (i == n - 1)
      kov = 2;
    else
      kov = sz[i + 1].first;
    int fi = sz[i].first, se = sz[i].second, si = p.size(), ut = p.back();
    if (fi == 3 && si <= 4 && kov == 3) {
      int q = sz[i + 1].second;
      pb(), add(se, p.back()), add(q, p.back()), pb(), add(se, p.back()), cnt++,
          add(se, cnt), add(q, cnt);
      p.push_back(cnt), p.push_back(ut), add(q, ut), i++;
    } else {
      if (si == 4 && kov == 4) kov--;
      fi--;
      while (fi > 0 && si + fi > kov + 1) add(se, p.back()), pb(), si--, fi--;
      add(se, p.back());
      while (fi > 0) cnt++, add(se, cnt), p.push_back(cnt), fi--;
      p.push_back(ut);
    }
  }
  cout << cnt << "\n";
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < sol[i].size(); j++) cout << sol[i][j] << " ";
    cout << "\n";
  }
  return 0;
}