#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

#define iter(i, n) for (int i = 1; i <= n; ++i)

const int NR = 2e5 + 100;

vector<int> mex[NR], ring;
int in[NR], n, p[NR], q[NR], sg[NR];

int calc(vector<int> &mex) {
	sort(mex.begin(), mex.end());
	int sg = 0;
	for (int i : mex) {
		if (i == sg) ++sg;
		else if (i > sg) break;
	}
	return sg;
}

int main() {
	scanf("%d", &n);
	iter(i, n) scanf("%d", &p[i]), ++in[p[i]];
	
	int h = 1, t = 0;
	iter(i, n) if (!in[i]) q[++t] = i;

	while (h <= t) {
		int i = q[h++];
		sg[i] = calc(mex[i]);
		mex[p[i]].push_back(sg[i]);
		if (!--in[p[i]]) q[++t] = p[i];
	}

	iter(i, n) if (in[i]) {
		for (int x = i; ; x = p[x]) {
			ring.push_back(x);
			sg[x] = calc(mex[x]);
			if (p[x] == i) break;
		}

		reverse(ring.begin(), ring.end());

		int len = ring.size(), lst = sg[ring[0]];
		for (int i = 1; i < len; ++i) {
			if (lst == sg[ring[i]]) ++lst;
			else lst = sg[ring[i]];
		}

		if (lst != sg[ring[0]]) puts("POSSIBLE");
		else {
			lst = sg[ring[0]] + 1;
			for (int i = 1; i < len; ++i) {
				if (lst == sg[ring[i]]) ++lst;
				else lst = sg[ring[i]];
			}
			if (lst == sg[ring[0]]) puts("POSSIBLE");
			else puts("IMPOSSIBLE");
		}
		break;
	}
	return 0;
}
### 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 <vector>
#include <algorithm>
using namespace std;

#define iter(i, n) for (int i = 1; i <= n; ++i)

const int NR = 2e5 + 100;

vector<int> mex[NR], ring;
int in[NR], n, p[NR], q[NR], sg[NR];

int calc(vector<int> &mex) {
	sort(mex.begin(), mex.end());
	int sg = 0;
	for (int i : mex) {
		if (i == sg) ++sg;
		else if (i > sg) break;
	}
	return sg;
}

int main() {
	scanf("%d", &n);
	iter(i, n) scanf("%d", &p[i]), ++in[p[i]];
	
	int h = 1, t = 0;
	iter(i, n) if (!in[i]) q[++t] = i;

	while (h <= t) {
		int i = q[h++];
		sg[i] = calc(mex[i]);
		mex[p[i]].push_back(sg[i]);
		if (!--in[p[i]]) q[++t] = p[i];
	}

	iter(i, n) if (in[i]) {
		for (int x = i; ; x = p[x]) {
			ring.push_back(x);
			sg[x] = calc(mex[x]);
			if (p[x] == i) break;
		}

		reverse(ring.begin(), ring.end());

		int len = ring.size(), lst = sg[ring[0]];
		for (int i = 1; i < len; ++i) {
			if (lst == sg[ring[i]]) ++lst;
			else lst = sg[ring[i]];
		}

		if (lst != sg[ring[0]]) puts("POSSIBLE");
		else {
			lst = sg[ring[0]] + 1;
			for (int i = 1; i < len; ++i) {
				if (lst == sg[ring[i]]) ++lst;
				else lst = sg[ring[i]];
			}
			if (lst == sg[ring[0]]) puts("POSSIBLE");
			else puts("IMPOSSIBLE");
		}
		break;
	}
	return 0;
}
```