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 (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#define N 510
#define NN N*N

int n;
int x[N];
int ans[NN];
int queue[NN];
int qfront, qback;

int main() {
  scanf("%d", &n);
  for(int i = 0; i < n; i++) scanf("%d", x+i);
  memset(ans, -1, sizeof(ans));
  for(int i = 0; i < n; i++) ans[x[i]-1] = i+1+1000;
  int now = 0;
  for(int i = 0; i < n*n; i++) {
    if(ans[i] != -1) {
      int u = ans[i]-1000;
      int cnt = 1;
      while(cnt < u) {
        if(ans[now] != -1) now++;
        else {ans[now++] = u; cnt++;}
      }
      if(now > i) {
        printf("No\n");
        return 0;
      }
    }
  }
  for(int i = 0; i < n*n; i++) {
    if(ans[i] == -1) {
      if(qfront == qback) {
        printf("No\n");
        return 0;
      }
      ans[i] = queue[qfront++];
    }
    if(ans[i] > 1000) {
      ans[i] -= 1000;
      for(int j = 0; j < n-ans[i]; j++) queue[qback++] = ans[i];
    }
  }

  printf("Yes\n");
  for(int i = 0; i < n*n; i++) {
    if(i) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
}