# N-1の長さは1本までしか使えない、N-2の長さは2本までしか使えない、、、、 # 1本でも減れば木にならないのでアウト # そして、gcd(A)=1でなければつながらない # 構築はUnion Findで愚直につながっていないところをつなげるか、時間間に合うか class UnionFind(): def __init__(self, n): self.n = n self.parents = [-1] * n def find(self, x): if self.parents[x] < 0: return x else: self.parents[x] = self.find(self.parents[x]) return self.parents[x] def unite(self, x, y): x = self.find(x) y = self.find(y) if x == y: return if self.parents[x] > self.parents[y]: x, y = y, x self.parents[x] += self.parents[y] self.parents[y] = x def size(self, x): return -self.parents[self.find(x)] def same(self, x, y): return self.find(x) == self.find(y) def members(self, x): root = self.find(x) return [i for i in range(self.n) if self.find(i) == root] def roots(self): return [i for i, x in enumerate(self.parents) if x < 0] def group_count(self): return len(self.roots()) def all_group_members(self): group_members = defaultdict(list) for member in range(self.n): group_members[self.find(member)].append(member) return group_members def __str__(self): return '\n'.join(f'{r}: {m}' for r, m in self.all_group_members().items()) N = int(input()) A = list(map(int, input().split())) from collections import Counter counted = Counter(A) length_test = True for d in range(N-1, 0, -1): if counted[d] > N-d: length_test = False import math import functools GCD = functools.reduce(math.gcd, A) if GCD == 1: GCD_test = True else: GCD_test = False def make_connection(): UF = UnionFind(N+1) ans_list = [] for a in A: for i in range(1, N+1): if UF.same(i, i+a) == False: UF.unite(i, i+a) ans_list.append(i) break for a in ans_list: print(a) if length_test == True and GCD_test == True: print('YES') make_connection() else: print('NO')