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 Maxn = 300000;
struct Node {
  int x, id;
  friend bool operator<(Node a, Node b) {
    if (a.x == b.x) {
      return a.id < b.id;
    }
    return a.x > b.x;
  }
} a[Maxn + 5];
deque<int> q;
vector<int> edge[Maxn + 5];
int main() {
  int f;
  scanf("%d", &f);
  for (int i = 1; i <= f; i++) {
    scanf("%d", &a[i].x);
    a[i].id = i;
  }
  sort(a + 1, a + 1 + f);
  int n = 0;
  for (int i = 1; i <= a[1].x; i++) {
    n++;
    q.push_back(n);
    edge[a[1].id].push_back(n);
  }
  for (int i = 2; i <= f; i++) {
    int e, k;
    e = a[i].x;
    k = a[i].id;
    int p = min(e - 1, ((int)q.size() + e - (i == f ? 3 : a[i + 1].x)) / 2);
    for (int j = 0; j < p - 1; j++) {
      edge[k].push_back(q.back());
      q.pop_back();
    }
    edge[k].push_back(q.back());
    for (int j = p; j < e - 1; j++) {
      n++;
      edge[k].push_back(n);
      q.push_back(n);
    }
    edge[k].push_back(q.front());
    q.push_back(q.front());
    q.pop_front();
  }
  printf("%d\n", n);
  for (int i = 1; i <= f; i++) {
    for (int j = 0; j < (int)edge[i].size(); j++) {
      printf("%d ", edge[i][j]);
    }
    puts("");
  }
  return 0;
}