#include <bits/stdc++.h>
using namespace std;
int n, m, K;
pair<int, int> a[100010];
vector<int> p[100010];
deque<int> q;
int main() {
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i].first), a[i].second = i;
  sort(a + 1, a + m + 1), reverse(a + 1, a + m + 1);
  for (int i = 1; i <= a[1].first; i++) {
    ++n;
    p[a[1].second].push_back(n);
    q.push_back(n);
  }
  for (int i = 2; i <= m; i++) {
    int E = a[i].first, id = a[i].second,
        r = min(E - 1, ((int)q.size() + E - (i == m ? 3 : a[i + 1].first)) / 2);
    for (int i = 1; i < r; i++) p[id].push_back(q.back()), q.pop_back();
    p[id].push_back(q.back());
    for (int i = r + 1; i < E; i++) ++n, p[id].push_back(n), q.push_back(n);
    p[id].push_back(q.front()), q.push_back(q.front()), q.pop_front();
  }
  printf("%d\n", n);
  for (int i = 1; i <= m; i++) {
    for (auto j : p[i]) printf("%d ", j);
    puts("");
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int n, m, K;
pair<int, int> a[100010];
vector<int> p[100010];
deque<int> q;
int main() {
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d", &a[i].first), a[i].second = i;
  sort(a + 1, a + m + 1), reverse(a + 1, a + m + 1);
  for (int i = 1; i <= a[1].first; i++) {
    ++n;
    p[a[1].second].push_back(n);
    q.push_back(n);
  }
  for (int i = 2; i <= m; i++) {
    int E = a[i].first, id = a[i].second,
        r = min(E - 1, ((int)q.size() + E - (i == m ? 3 : a[i + 1].first)) / 2);
    for (int i = 1; i < r; i++) p[id].push_back(q.back()), q.pop_back();
    p[id].push_back(q.back());
    for (int i = r + 1; i < E; i++) ++n, p[id].push_back(n), q.push_back(n);
    p[id].push_back(q.front()), q.push_back(q.front()), q.pop_front();
  }
  printf("%d\n", n);
  for (int i = 1; i <= m; i++) {
    for (auto j : p[i]) printf("%d ", j);
    puts("");
  }
}

```