#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n, Q[201000], head, tail, Deg[201000], v[201000], D[201000], E[201000];
vector<int> F[201000];
bool Pos(int a, int d) {
	D[a] = d;
	int x = a, i;
	while (1) {
		x = E[x];
		for (i = 0; i < F[x].size(); i++) {
			v[D[F[x][i]]] = -x;
		}
		for (i = 0;; i++)if (v[i] != -x)break;
		if (x == a && i != d)return false;
		D[x] = i;
		if (x == a)return true;
	}
}
int main() {
	int i, x;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d", &x);
		E[i] = x;
		F[x].push_back(i);
		Deg[x]++;
	}
	for (i = 1; i <= n; i++) {
		if (!Deg[i])Q[++tail] = i;
	}
	while (head < tail) {
		x = Q[++head];
		for (i = 0; i < F[x].size(); i++) {
			v[D[F[x][i]]] = x;
		}
		for (i = 0;; i++)if (v[i] != x)break;
		D[x] = i;
		int y = E[x];
		Deg[y]--;
		if (!Deg[y]) Q[++tail] = y;
	}
	for (i = 1; i <= n; i++)if (Deg[i])break;
	x = i;
	for (i = 0; i < F[x].size(); i++)if(!Deg[F[x][i]])v[D[F[x][i]]] = x;
	for (i = 0;; i++)if (v[i] != x)break;
	int r1 = i;
	for (i = i + 1;; i++)if (v[i] != x)break;
	int r2 = i;
	if (Pos(x, r1))puts("POSSIBLE");
	else {
		if (Pos(x, r2))puts("POSSIBLE");
		else puts("IMPOSSIBLE");
	}
}
### Prompt

Your task is to create a cpp solution to 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<algorithm>
#include<vector>
using namespace std;
int n, Q[201000], head, tail, Deg[201000], v[201000], D[201000], E[201000];
vector<int> F[201000];
bool Pos(int a, int d) {
	D[a] = d;
	int x = a, i;
	while (1) {
		x = E[x];
		for (i = 0; i < F[x].size(); i++) {
			v[D[F[x][i]]] = -x;
		}
		for (i = 0;; i++)if (v[i] != -x)break;
		if (x == a && i != d)return false;
		D[x] = i;
		if (x == a)return true;
	}
}
int main() {
	int i, x;
	scanf("%d", &n);
	for (i = 1; i <= n; i++) {
		scanf("%d", &x);
		E[i] = x;
		F[x].push_back(i);
		Deg[x]++;
	}
	for (i = 1; i <= n; i++) {
		if (!Deg[i])Q[++tail] = i;
	}
	while (head < tail) {
		x = Q[++head];
		for (i = 0; i < F[x].size(); i++) {
			v[D[F[x][i]]] = x;
		}
		for (i = 0;; i++)if (v[i] != x)break;
		D[x] = i;
		int y = E[x];
		Deg[y]--;
		if (!Deg[y]) Q[++tail] = y;
	}
	for (i = 1; i <= n; i++)if (Deg[i])break;
	x = i;
	for (i = 0; i < F[x].size(); i++)if(!Deg[F[x][i]])v[D[F[x][i]]] = x;
	for (i = 0;; i++)if (v[i] != x)break;
	int r1 = i;
	for (i = i + 1;; i++)if (v[i] != x)break;
	int r2 = i;
	if (Pos(x, r1))puts("POSSIBLE");
	else {
		if (Pos(x, r2))puts("POSSIBLE");
		else puts("IMPOSSIBLE");
	}
}
```