#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <sstream>
#include <algorithm>
#pragma comment(linker, "/STACK:16000000")
using namespace std;

const int Maxn = 200005;
const int Maxm = 100;

bool has[Maxn][Maxm];
int res[Maxn];
bool tk[Maxn];
int root, rootval;
int n;
vector <int> neigh[Maxn];
int P[Maxn];

void Solve(int v)
{
	for (int i = 0; i < neigh[v].size(); i++) {
		int u = neigh[v][i];
		if (u == root) has[v][rootval] = true;
		else { Solve(u); has[v][res[u]] = true; }
	}
	res[v] = 0;
	while (has[v][res[v]]) res[v]++;
	for (int i = 0; i < neigh[v].size(); i++) {
		int u = neigh[v][i];
		if (u == root) has[v][rootval] = false;
		else has[v][res[u]] = false;
	}
}

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &P[i]); 
		neigh[P[i]].push_back(i);
	}
	root = 1;
	while (!tk[root]) {
		tk[root] = true; root = P[root];
	}
	for (rootval = 0; rootval < Maxm; rootval++) {
		res[root] = rootval;
		for (int j = 0; j < neigh[root].size(); j++) {
			int u = neigh[root][j]; Solve(u);
			has[root][res[u]] = true;
		}
		int my = 0;
		while (has[root][my]) my++;
		if (res[root] == my) { printf("POSSIBLE\n"); return 0; }
		for (int j = 0; j < neigh[root].size(); j++) {
			int u = neigh[root][j];
			has[root][res[u]] = false;
		}
	}
	printf("IMPOSSIBLE\n");
	return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

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 <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstring>
#include <sstream>
#include <algorithm>
#pragma comment(linker, "/STACK:16000000")
using namespace std;

const int Maxn = 200005;
const int Maxm = 100;

bool has[Maxn][Maxm];
int res[Maxn];
bool tk[Maxn];
int root, rootval;
int n;
vector <int> neigh[Maxn];
int P[Maxn];

void Solve(int v)
{
	for (int i = 0; i < neigh[v].size(); i++) {
		int u = neigh[v][i];
		if (u == root) has[v][rootval] = true;
		else { Solve(u); has[v][res[u]] = true; }
	}
	res[v] = 0;
	while (has[v][res[v]]) res[v]++;
	for (int i = 0; i < neigh[v].size(); i++) {
		int u = neigh[v][i];
		if (u == root) has[v][rootval] = false;
		else has[v][res[u]] = false;
	}
}

int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &P[i]); 
		neigh[P[i]].push_back(i);
	}
	root = 1;
	while (!tk[root]) {
		tk[root] = true; root = P[root];
	}
	for (rootval = 0; rootval < Maxm; rootval++) {
		res[root] = rootval;
		for (int j = 0; j < neigh[root].size(); j++) {
			int u = neigh[root][j]; Solve(u);
			has[root][res[u]] = true;
		}
		int my = 0;
		while (has[root][my]) my++;
		if (res[root] == my) { printf("POSSIBLE\n"); return 0; }
		for (int j = 0; j < neigh[root].size(); j++) {
			int u = neigh[root][j];
			has[root][res[u]] = false;
		}
	}
	printf("IMPOSSIBLE\n");
	return 0;
}
```