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 Python (3.4.3) (judge verdict: Wrong Answer)

def main():
    n = int(input())
    X = list(map(int, input().split()))

    ans = [-1]*(n**2)
    left = []
    right = []
    for idx, x in enumerate(X):
        idx += 1
        ans[x-1] = idx
        left.append((x, idx-1, idx))
        right.append((x, n-idx, idx))

    left.sort(key=lambda x: -x[0])
    right.sort(key=lambda x: x[0])

    amt, cnt = 0, 0

    for i in range(n**2):
        if ans[i] != -1:
            continue
        while amt == 0 and cnt < n:
            cnt += 1
            x, amt, idx = left.pop()
        if amt == 0:
            break
        if x < i:
            print("No")
            return
        amt -= 1
        ans[i] = idx

    amt, cnt = 0, 0

    for i in range(n**2-1, -1, -1):
        if ans[i] != -1:
            continue
        while amt == 0 and cnt < n:
            cnt += 1
            x, amt, idx = right.pop()
        if amt == 0:
            break
        if x > i:
            print("No")
            return
        amt -= 1
        ans[i] = idx
    print("Yes")
    print(*ans)


main()