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
#include <bits/stdc++.h>
using namespace std;

const int MN = 501;
int a[MN * MN], mx[MN], ptr, lst[MN];
void add(int v, bool b) {
	while (a[ptr])
		ptr++;
//	cerr << v << 'p' << ptr << 'b' << b << 'l' << lst[v] << endl;
	a[ptr] = v;
	if (!b) {
		if (ptr > lst[v])
			cout << "No\n", exit(0);
	} else
		if (ptr < lst[v])
			cout << "No\n", exit(0);
	ptr++;
}

int main() {
	int n;
	cin >> n;
	pair <int, int> x[MN];
	for (int i = 0; i < n; i++) {
		cin >> x[i].first;
		a[--x[i].first] = i + 1;
		x[i].second = i + 1;
		lst[i + 1] = x[i].first;
	}
	sort(x, x + n);
	for (int i = 0; i < n; i++) 
		for (int j = 1; j < x[i].second; j++)
			add(x[i].second, false);
	for (int i = 0; i < n; i++) 
		for (int j = x[i].second; j < n; j++)
			add(x[i].second, true);
	cout << "Yes" << endl;
	for (int i = 0; i < n * n; i++)
		cout << a[i] << endl;
}
