#include <bits/stdc++.h>
 
using namespace std;
 
int n, p[200005], in[200005], ta, tb;
set< int > s[200005];
vector< int > x, y;
bool v[200005];
 
void go(int a) {
	v[a] = 1;
	for (int i = 0; ; i++)
		if (s[a].find(i) == s[a].end()) {
			s[p[a]].insert(i);
			break;
		}
	in[p[a]]--;
	if (in[p[a]] == 0)
		go(p[a]);
}
 
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", p + i);
		in[p[i]]++;
	}
	for (int i = 1; i <= n; i++)
		if (!in[i] && !v[i])
			go(i);
	for (int i = 1; i <= n; i++)
		if (!v[i]) {
			ta = i;
			while (1) {
				x.push_back(ta);
				if (p[ta] == i)
					break;
				ta = p[ta];
			}
			break;
		}
	for (int i = 0; y.size() < 2; i++)
		if (s[x[0]].find(i) == s[x[0]].end())
			y.push_back(i);
	for (int i = 0; i < 2; i++) {
		in[0] = y[i];
		for (int j = 1; j < x.size(); j++)
			for (int k = 0; ; k++)
				if (in[j - 1] != k && s[x[j]].find(k) == s[x[j]].end()) {
					in[j] = k;
					break;
				}
		if (i == 0 && in[x.size() - 1] != y[0]) {
			printf("POSSIBLE\n");
			return 0;
		}
		if (i == 1 && in[x.size() - 1] == y[0]) {
			printf("POSSIBLE\n");
			return 0;
		}
	}
	printf("IMPOSSIBLE\n");
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
 
using namespace std;
 
int n, p[200005], in[200005], ta, tb;
set< int > s[200005];
vector< int > x, y;
bool v[200005];
 
void go(int a) {
	v[a] = 1;
	for (int i = 0; ; i++)
		if (s[a].find(i) == s[a].end()) {
			s[p[a]].insert(i);
			break;
		}
	in[p[a]]--;
	if (in[p[a]] == 0)
		go(p[a]);
}
 
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", p + i);
		in[p[i]]++;
	}
	for (int i = 1; i <= n; i++)
		if (!in[i] && !v[i])
			go(i);
	for (int i = 1; i <= n; i++)
		if (!v[i]) {
			ta = i;
			while (1) {
				x.push_back(ta);
				if (p[ta] == i)
					break;
				ta = p[ta];
			}
			break;
		}
	for (int i = 0; y.size() < 2; i++)
		if (s[x[0]].find(i) == s[x[0]].end())
			y.push_back(i);
	for (int i = 0; i < 2; i++) {
		in[0] = y[i];
		for (int j = 1; j < x.size(); j++)
			for (int k = 0; ; k++)
				if (in[j - 1] != k && s[x[j]].find(k) == s[x[j]].end()) {
					in[j] = k;
					break;
				}
		if (i == 0 && in[x.size() - 1] != y[0]) {
			printf("POSSIBLE\n");
			return 0;
		}
		if (i == 1 && in[x.size() - 1] == y[0]) {
			printf("POSSIBLE\n");
			return 0;
		}
	}
	printf("IMPOSSIBLE\n");
	return 0;
}
```