There is a directed graph with N vertices and N edges. The vertices are numbered 1, 2, ..., N.

The graph has the following N edges: (p_1, 1), (p_2, 2), ..., (p_N, N), and the graph is weakly connected. Here, an edge from Vertex u to Vertex v is denoted by (u, v), and a weakly connected graph is a graph which would be connected if each edge was bidirectional.

We would like to assign a value to each of the vertices in this graph so that the following conditions are satisfied. Here, a_i is the value assigned to Vertex i.

* Each a_i is a non-negative integer.
* For each edge (i, j), a_i \neq a_j holds.
* For each i and each integer x(0 ≤ x < a_i), there exists a vertex j such that the edge (i, j) exists and x = a_j holds.



Determine whether there exists such an assignment.

Constraints

* 2 ≤ N ≤ 200 000
* 1 ≤ p_i ≤ N
* p_i \neq i
* The graph is weakly connected.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

If the assignment is possible, print `POSSIBLE`; otherwise, print `IMPOSSIBLE`.

Examples

Input

4
2 3 4 1


Output

POSSIBLE


Input

3
2 3 1


Output

IMPOSSIBLE


Input

4
2 3 1 1


Output

POSSIBLE


Input

6
4 5 6 5 6 4


Output

IMPOSSIBLE
# 結局は木+1辺
from collections import defaultdict
import heapq as hq

N = int(input())
P = list(map(lambda x:int(x)-1,input().split()))
D = [0]*N
for p in P:
  D[p] += 1

child_L = defaultdict(list)

S = [p for p in range(N) if D[p] == 0]
L = [None]*N

while S:
  n = S.pop()

  q = child_L[n]
  del child_L[n]

  hq.heapify(q)
  i = 0
  while q:
    t = hq.heappop(q)
    if i < t:
      break
    else:
      i += (i==t)
  
  L[n] = i
  m = P[n]
  child_L[m].append(i)
  D[m] -= 1
  if D[m] == 0:
    S.append(m)

# cycle check
try:
  start = D.index(1)
except ValueError:
  print('POSSIBLE')
  exit()

def helper(n):
  q = child_L[n]
  del child_L[n]
  
  hq.heapify(q)
  i = 0
  while q:
    t = hq.heappop(q)
    if i < t:
      break
    else:
      i += (i==t)

  j = i+1
  while q:
    t = hq.heappop(q)
    if j < t:
      break
    else:
      j += (j==t)

  return (i,j)


s1,s2 = helper(start)
G = []
n = P[start]
while n != start:
  G.append(helper(n))
  n = P[n]


del N,P,D,child_L,S,L


# 可能な初期値をそれぞれシミュレート
# 1
n = s1
for g in G:
  if g[0] == n:
    n = g[1]
  else:
    n = g[0]
if n != s1:
  print('POSSIBLE')
  exit()

# 2
n = s2
for g in G:
  if g[0] == n:
    n = g[1]
  else:
    n = g[0]
if n == s1:
  print('POSSIBLE')
  exit()

print('IMPOSSIBLE')