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
#include <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, P[200000], a[200000];
bool vis[200000];
vector<int>rev[200000];

void dfs(int x) {
	if (a[x] != -1)return;
	set<int>st;
	for (int i : rev[x]) {
		dfs(i);
		st.insert(a[i]);
	}
	for (int i = 0;; ++i) {
		if (st.find(i) == st.end()) {
			a[x] = i;
			return;
		}
	}
}

int main() {
	cin >> N;
	for (int i = 0; i < N; ++i) {
		cin >> P[i];
		--P[i];
		rev[P[i]].push_back(i);
		a[i] = -1;
	}
	int p, q;
	p = 0;
	while (!vis[P[p]]) {
		p = P[p];
		vis[p] = true;
	}
	q = P[p];
	set<int>st;
	for (int i : rev[q]) {
		if (i == p)continue;
		dfs(i);
		st.insert(a[i]);
	}
	int t;
	for (int i = 0;; ++i) {
		if (st.find(i) == st.end()) {
			t = a[q] = i;
			break;
		}
	}
	dfs(p);
	if (a[q] != a[p]) {
		puts("POSSIBLE");
		return 0;
	}
	memset(a, -1, sizeof a);
	for (int i : rev[q]) {
		if (i == p)continue;
		dfs(i);
	}
	st.insert(t);
	for (int i = 0;; ++i) {
		if (st.find(i) == st.end()) {
			a[q] = i;
			break;
		}
	}
	dfs(p);
	if (a[p] == t) {
		puts("POSSIBLE");
		return 0;
	}
	puts("IMPOSSIBLE");
	return 0;
}
