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
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

"""
・pseudo-tree graph
・in_deg = 1が保証されているので、向きの揃ったcycleがある
・1箇所の2択を決めれば全部決まる
"""

N = int(readline())
parent = [0] + list(map(int,read().split()))

child = [[] for _ in range(N+1)]
for i,x in enumerate(parent):
    child[x].append(i)

out_deg = [len(x) for x in child]

out_deg

G = [-1] * (N+1)

# とりあえず唯一のサイクル以外を処理：out_deg=0の頂点を捨てていく

stack = [i for i,x in enumerate(out_deg) if not x]
while stack:
    x = stack.pop()
    se = set(G[c] for c in child[x])
    g = 0
    while g in se:
        g += 1
    G[x] = g
    p = parent[x]
    out_deg[p] -= 1
    if not out_deg[p]:
        stack.append(p)

"""
・grundy数の候補は2種。
・片方決め打って計算。2周grundy数を計算してstableになっていればよい。
"""

for i,x in enumerate(out_deg[1:],1):
    if x==1:
        root = i
        break

cycle_len = sum(out_deg[1:])

is_stable = False
x = root
for _ in range(cycle_len*2+10):
    se = set(G[c] for c in child[x])
    g = 0
    while g in se:
        g += 1
    if g == G[x]:
        is_stable = True
        break
    G[x] = g
    x = parent[x]

answer = 'POSSIBLE' if is_stable else 'IMPOSSIBLE'
print(answer)