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>

typedef std::pair<int, int> pi;
std::priority_queue<pi, std::vector<pi>, std::greater<pi> > q;
const int MAXN = 510 * 510;
int n, A[MAXN], B[MAXN];
std::vector<pi> add[MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1, t; i <= n; ++i) {
		std::cin >> t; B[i] = t;
		for (int j = 1; j < i; ++j) add[1].emplace_back(t - 1, i);
		add[t].emplace_back(t, i);
		for (int j = i + 1; j <= n; ++j) add[t + 1].emplace_back(n * n, i);
	}
	for (int i = 1; i <= n * n; ++i) {
		for (auto t : add[i]) q.push(t);
		if (q.empty() || q.top().first < i)
			return std::cout << "No" << std::endl, 0;
		A[i] = q.top().second; q.pop();
	}
	std::cout << "Yes" << std::endl;
	for (int i = 1; i <= n * n; ++i)
		std::cout << A[i] << (" \n" [i == n * n]);
	return 0;
}
