Problem: Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root. 

Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.

Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.

The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.

Input

The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.

The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.

Output

If there is only one tree matching this sequence, print "perfect".

Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.

These treese should be non-isomorphic and should match the given sequence.

Examples

Input

2
1 1 1


Output

perfect


Input

2
1 2 2


Output

ambiguous
0 1 1 3 3
0 1 1 3 2

Note

The only tree in the first example and the two printed trees from the second example are shown on the picture:

<image>

Solution (in PYTHON3):

Solution:
"""
    Author - Satwik Tiwari .
    21th NOV , 2020  - Saturday
"""

#===============================================================================================
#importing some useful libraries.

from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key

# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt

from copy import deepcopy
from collections import deque


from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
    for pp in range(t):
        solve(pp)
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    y%=(p-1)  #not so sure about this. used when y>p-1. if p is prime.
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))



def solve(case):
    n = int(inp())
    a = lis()

    cnt = False
    for i in range(1,n+1):
        if(a[i] > 1 and a[i-1] > 1):
            cnt = True

    if(not cnt):
        print('perfect')
        return


    print('ambiguous')
    ans = [0]

    par = 1
    for i in range(1,n+1):
        # print(i,par)
        for j in range(a[i]):
            ans.append(par)
        par+=a[i-1]

    # print(ans)

    ans2 = [0]
    par = 1
    for i in range(1,n+1):
        # print(i,a[i-1],par)
        for j in range(a[i]-1):
            ans2.append(par)
        if(a[i-1] > 1):
            ans2.append(par+1)
        else:
            ans2.append(par)
        par+=a[i-1]


    # print(ans2)

    print(' '.join(str(ans[i]) for i in range(len(ans))))
    print(' '.join(str(ans2[i]) for i in range(len(ans2))))




testcase(1)
# testcase(int(inp()))