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;
const int MAX_N = 100000;
int n;
int a[MAX_N + 9];
int id[MAX_N + 9];
int ans1;
vector<int> ans2[MAX_N + 9];
vector<int> peri;
bool Cmp(int x, int y) { return a[x] > a[y]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    id[i] = i;
  }
  a[n + 1] = 3;
  id[n + 1] = n + 1;
  sort(id + 1, id + n + 1, Cmp);
  ans1 = a[id[1]];
  for (int i = 1; i <= a[id[1]]; i++) {
    peri.push_back(i);
    ans2[id[1]].push_back(i);
  }
  for (int i = 2; i <= n; i++) {
    int x = id[i];
    int xx = id[i + 1];
    if (peri.size() == 4 && a[x] == 3) {
      ans2[x].push_back(peri[1]);
      ans2[x].push_back(peri[2]);
      ans2[x].push_back(peri[3]);
      peri.erase(peri.begin() + 2);
    } else {
      int d = min(((int)peri.size() + a[x] - a[xx]) / 2, a[x] - 1);
      ans2[x].push_back(peri[0]);
      ans2[x].push_back(peri.back());
      for (int j = 3; j <= d + 1; j++) {
        peri.pop_back();
        ans2[x].push_back(peri.back());
      }
      for (int j = d + 2; j <= a[x]; j++) {
        ans1++;
        peri.push_back(ans1);
        ans2[x].push_back(ans1);
      }
    }
  }
  cout << ans1 << "\n";
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < a[i]; j++) {
      cout << ans2[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}