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 N = 3e5 + 1;
struct Face {
  vector<int> Ver;
  int siz;
  int idx;
  int &operator[](int i) { return Ver[i]; }
} F[N];
int C[N];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int s;
    cin >> s;
    F[i].Ver.resize(s);
    F[i].siz = s;
    F[i].idx = i;
  }
  sort(F, F + n, [&](const Face &a, const Face &b) { return a.siz > b.siz; });
  for (int i = n - 1; i; --i) C[i] = C[i + 1] + F[i].siz - 2;
  iota(F[0].Ver.begin(), F[0].Ver.end(), 1);
  deque<int> Combi;
  for (int x : F[0].Ver) Combi.push_back(x);
  int tot = Combi.size();
  for (int i = 1; i < n; ++i) {
    if (Combi.size() + F[i].siz - 4 < C[i + 1]) {
      int ver = Combi.back();
      Combi.pop_back();
      F[i][0] = Combi.back();
      Combi.pop_back();
      F[i].Ver.back() = ver;
      for (int j = 1; j < F[i].siz - 1; ++j) F[i][j] = ++tot;
      for (int x : F[i].Ver) Combi.push_back(x);
    } else {
      int k = (Combi.size() + F[i].siz - C[i + 1]) / 2;
      if (k >= Combi.size()) k = Combi.size() - 1;
      if (k > F[i].siz) k = F[i].siz;
      for (int j = 0; j < k; ++j) {
        assert(Combi.size());
        F[i][j] = Combi.back();
        if (j < k - 1) Combi.pop_back();
      }
      for (int j = k; j < F[i].siz; ++j) {
        F[i][j] = ++tot;
        Combi.push_back(tot);
      }
      Combi.push_back(F[i][0]);
    }
    Combi.push_front(Combi.back());
    Combi.pop_back();
  }
  sort(F, F + n, [&](const Face &a, const Face &b) { return a.idx < b.idx; });
  cout << tot << "\n";
  for (int i = 0; i < n; ++i) {
    for (int x : F[i].Ver) cout << x << " ";
    cout << "\n";
  }
}