#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n;
array<int, 2> a[N];
vector<int> v[N];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i][0]), a[i][1] = i;
  sort(a + 1, a + 1 + n), reverse(a + 1, a + 1 + n);
  deque<int> q;
  int ind = 0;
  for (int i = (1); i <= (a[1][0]); ++i) {
    q.push_back(++ind);
    v[a[1][1]].push_back(ind);
  }
  for (int i = (2); i <= (n); ++i) {
    int w = min(a[i][0] - 1,
                (((int)(q).size()) + a[i][0] - (i == n ? 3 : a[i + 1][0])) / 2);
    for (int j = (1); j <= (w - 1); ++j) {
      v[a[i][1]].push_back(q.back());
      q.pop_back();
    }
    v[a[i][1]].push_back(q.back());
    for (int j = (w + 1); j <= (a[i][0] - 1); ++j) {
      v[a[i][1]].push_back(++ind);
      q.push_back(ind);
    }
    v[a[i][1]].push_back(q.front());
    q.push_back(q.front());
    q.pop_front();
  }
  printf("%d\n", ind);
  for (int i = (1); i <= (n); ++i) {
    for (auto x : v[i]) printf("%d ", x);
    puts("");
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n;
array<int, 2> a[N];
vector<int> v[N];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%d", &a[i][0]), a[i][1] = i;
  sort(a + 1, a + 1 + n), reverse(a + 1, a + 1 + n);
  deque<int> q;
  int ind = 0;
  for (int i = (1); i <= (a[1][0]); ++i) {
    q.push_back(++ind);
    v[a[1][1]].push_back(ind);
  }
  for (int i = (2); i <= (n); ++i) {
    int w = min(a[i][0] - 1,
                (((int)(q).size()) + a[i][0] - (i == n ? 3 : a[i + 1][0])) / 2);
    for (int j = (1); j <= (w - 1); ++j) {
      v[a[i][1]].push_back(q.back());
      q.pop_back();
    }
    v[a[i][1]].push_back(q.back());
    for (int j = (w + 1); j <= (a[i][0] - 1); ++j) {
      v[a[i][1]].push_back(++ind);
      q.push_back(ind);
    }
    v[a[i][1]].push_back(q.front());
    q.push_back(q.front());
    q.pop_front();
  }
  printf("%d\n", ind);
  for (int i = (1); i <= (n); ++i) {
    for (auto x : v[i]) printf("%d ", x);
    puts("");
  }
  return 0;
}

```