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
import os
import sys
from collections import deque, defaultdict

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353

N = int(sys.stdin.buffer.readline())
X = list(map(int, sys.stdin.buffer.readline().split()))

ans = [None] * (N ** 2 + 1)
# 左にあるものから使う
P = list(sorted([(pos, i) for i, pos in enumerate(X, 1)]))

que = deque()
for p, i in P:
    ans[p] = i
    for _ in range(i - 1):
        que.append(i)

for p in range(1, N ** 2 + 1):
    if ans[p] is not None:
        i = ans[p]
        for _ in range(N - i):
            que.append(i)
    else:
        if not que:
            break
        ans[p] = que.popleft()


def is_ok(ans, que):
    if que:
        return False
    for a in ans[1:]:
        if a is None:
            return False
    counts = defaultdict(int)
    C = [0]
    for a in ans[1:]:
        counts[a] += 1
        C.append(counts[a])
    for i, x in enumerate(X, 1):
        if C[x] != i:
            return False
    return True


if is_ok(ans, que):
    print('Yes')
    print(*ans[1:])
else:
    print('No')
