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_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
  }
  if (n == 1) {
    cout << a[0].first << endl;
    for (int i = 1; i <= a[0].first; i++) cout << i << ' ';
    return 0;
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  a.push_back(make_pair(2, 0));
  vector<vector<int>> face(n);
  vector<int> current;
  int cur = 1;
  int center = 1;
  face[0].push_back(1);
  for (int i = 2; i <= a[0].first; i++) {
    cur++;
    face[0].push_back(i);
    current.push_back(i);
  }
  for (int i = 1; i < n; i++) {
    int sz1 = current.size();
    int sz2 = a[i].first;
    if (sz1 == 2) {
      face[i].push_back(center);
      face[i].push_back(current[1]);
      cur++;
      face[i].push_back(cur);
      current.push_back(cur);
      continue;
    }
    int need = max(sz1 - sz2 + 2, a[i + 1].first);
    need = min(need, sz1 + sz2 - 4);
    if (need % 2 != (sz1 + sz2) % 2) need++;
    int common = (sz1 + sz2 + 2 - need) / 2;
    for (int j = 0; j < common; j++) {
      face[i].push_back(current.back());
      current.pop_back();
    }
    current.push_back((face[i].back()));
    for (int j = 0; j < sz2 - common; j++) {
      cur++;
      face[i].push_back(cur);
      current.push_back(cur);
    }
    current.push_back(face[i][0]);
  }
  cout << cur << endl;
  vector<vector<int>> kek(n);
  for (int i = 0; i < n; i++) kek[a[i].second] = face[i];
  for (int i = 0; i < n; i++) {
    for (auto it : kek[i]) cout << it << ' ';
    cout << endl;
  }
}