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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000;
const long long LINF = INF * (long long)INF;
const int MAX = 300007;
const int MOD = 1000000007;
const double Pi = acos(-1.0);
vector<int> res[MAX];
int main(int argc, char* argv[]) {
  int n;
  cin >> n;
  vector<pair<int, int> > A;
  for (int i = (0); i < (n); i++) {
    int x;
    cin >> x;
    A.push_back(make_pair(x, i));
  }
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  int s = 0;
  for (int i = (1); i < ((int)A.size()); i++) {
    s += A[i].first;
  }
  int cnt = A[0].first;
  vector<int> C;
  for (int i = (0); i < (cnt); i++) {
    C.push_back(i);
  }
  res[A[0].second] = C;
  for (int i = (1); i < (n); i++) {
    int c = 1;
    int nx = (i + 1 < n ? A[i + 1].first : 0);
    while (c < A[i].first - 1 && c < (int)C.size() - 2 &&
           (int)C.size() + A[i].first - 2 * (c + 1) >= nx) {
      ++c;
    }
    int id = A[i].second;
    int nv = A[i].first - 1 - c;
    for (int j = (0); j < (c); j++) {
      res[id].push_back(C.back());
      if (j + 1 < c) C.pop_back();
    }
    for (int j = (0); j < (nv); j++) {
      res[id].push_back(cnt);
      C.push_back(cnt);
      ++cnt;
    }
    res[id].push_back(C[0]);
    if ((int)C.size() <= 4) {
      C.push_back(C[0]);
      C.erase(C.begin());
    }
  }
  cout << cnt << endl;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < ((int)res[i].size()); j++) {
      cout << res[i][j] + 1 << ' ';
    }
    cout << endl;
  }
  cerr << (int)C.size() << endl;
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << endl;
}