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;
struct Node {
  int x, id;
  bool operator<(const Node &b) const { return x > b.x; }
} a[100000 + 5];
vector<int> ans[100000 + 5];
int que[100000 * 4 + 5], S, T, ans1;
void sew(int x, int si, int si1) {
  for (int i = 1; i < si; i++) ans[x].push_back(que[--T]);
  ans[x].push_back(que[T - 1]);
  for (int i = si + 1; i < si1; i++) ans[x].push_back(que[T++] = ++ans1);
  ans[x].push_back(que[T++] = que[S++]);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i), a[i].id = i;
  sort(a, a + n);
  a[n] = {3, 0};
  for (int i = 0; i < a[0].x; i++)
    que[T++] = ++ans1, ans[a[0].id].push_back(ans1);
  for (int i = 1; i < n; i++) {
    int w = a[i].x, si = min((T - S + w - a[i + 1].x) / 2, w - 1);
    sew(a[i].id, si, w);
  }
  printf("%d\n", ans1);
  for (int i = 0; i < n; i++) {
    for (int j : ans[i]) printf("%d ", j);
    puts("");
  }
}