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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>

using namespace std;

const int N = 555;

int n, x[N], a[N * N];
int ptr = 1;

int go(int times, int who) {
	int first = -1;
	for (int i = 0; i < times; i++) {
		while (a[ptr] != -1) ptr++;
		a[ptr] = who;
		if (first == -1) first = ptr;
	}
	return first;
}

int main() {
	scanf("%d", &n);
	vector<pair<int, int> > pos;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &x[i]);
		pos.push_back({x[i], i});
	}
	fill(a + 1, a + n * n + 1, -1);
	sort(pos.begin(), pos.end());
	for (auto p : pos) {
		int where = p.first, who = p.second;
		go(who - 1, who);
		if (a[where] != -1) {
			// printf("%d %d\n", where, who);
			puts("No");
			return 0;
		}
		a[where] = who;
	}
	for (auto p : pos) {
		int where = p.first, who = p.second;
		if (n - who && go(n - who, who) < where) {
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	for (int i = 1; i <= n * n; i++) {
		printf("%d%c", a[i], "\n "[i < n * n]);
	}
	return 0;
}
