Problem: You are given an integer sequence x of length N. Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

* a is N^2 in length, containing N copies of each of the integers 1, 2, ..., N.
* For each 1 ≤ i ≤ N, the i-th occurrence of the integer i from the left in a is the x_i-th element of a from the left.

Constraints

* 1 ≤ N ≤ 500
* 1 ≤ x_i ≤ N^2
* All x_i are distinct.

Input

The input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N


Output

If there does not exist an integer sequence a that satisfies all the conditions, print `No`. If there does exist such an sequence a, print `Yes` in the first line, then print an instance of a in the second line, with spaces inbetween.

Examples

Input

3
1 5 9


Output

Yes
1 1 1 2 2 2 3 3 3


Input

2
4 1


Output

No

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef pair<int,int> P;

int main(){
	int n,x[500],a[500 * 500] = {};
	cin >> n;
	vector<P> vec;
	for(int i = 0;i < n;i++) {
		cin >> x[i];
		vec.push_back(P(x[i] - 1,i + 1));
	}
	sort(vec.begin(),vec.end());
	int pos = 0;
	for(int i = 0;i < n;i++){
		int index = vec[i].first,v = vec[i].second;
		for(int j = 0;j < v - 1;j++){
			while(pos < index && a[pos]) pos++;
			if(pos >= index){
				cout << "No" << endl;
				return 0;
			}
			a[pos] = v;
		}
		a[index] = v;
	}
	reverse(vec.begin(),vec.end());
	pos = n * n - 1;
	for(int i = 0;i < n;i++){
		int index = vec[i].first,v = vec[i].second;
		for(int j = 0;j < n - v;j++){
			while(pos > index && a[pos]) pos--;
			if(pos <= index){
				cout << "No" << endl;
				return 0;
			}
			a[pos] = v;
		}
	}
	cout << "Yes" << endl;
	for(int i = 0;i < n * n;i++) cout << (i ? " " : "") << a[i];
	cout << endl;
	return 0;
}