Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

3
1 5 9

Sample Output 1

Yes
1 1 1 2 2 2 3 3 3

For example, the second occurrence of the integer 2 from the left in a in the output is the fifth element of a from the left.
Similarly, the condition is satisfied for the integers 1 and 3.

Sample Input 2

2
4 1

Sample Output 2

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
using namespace std;

const int MAXN = 500 * 500 + 10;
int N;
int A[MAXN];
int x[MAXN], rk[MAXN];
int cnt[MAXN];

int main() {
	register int i, j = 0;
	scanf("%d", &N);
	for(i = 1; i <= N; ++i)
		scanf("%d", x + i), rk[i] = i, cnt[i] = N;
	sort(rk + 1, rk + N + 1, [&](int a, int b) {return x[a] < x[b];});
	for(i = 1; i <= N; ++i) {
		while(cnt[rk[i]] > N - rk[i] + 1) {
			--cnt[rk[i]];
			while(A[j + 1]) ++j;
			A[++j] = rk[i];
		}
		if(A[x[rk[i]]]) return puts("No"), 0;
		A[x[rk[i]]] = rk[i]; --cnt[rk[i]];
	}
	i = 1;
	for(; j <= N * N; ++j) {
		if(A[j]) continue;
		if(cnt[rk[i]] > 0) --cnt[rk[i]], A[j] = rk[i];
		else {while(cnt[rk[i]] == 0) ++i; --cnt[rk[i]]; A[j] = rk[i];}
		if(j <= x[rk[i]]) return puts("No"), 0;
	}
	puts("Yes");
	for(i = 1; i <= N * N; ++i)
		printf("%d%c", A[i], " \n"[i == N * N]);
	return 0;
}