#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int to[200000];
vector<int>frm[200000];
bool flag[220000];
bool iscirc[200000];
typedef pair<int, int>pii;
pii dfs(int node)
{
	vector<int>v;
	for (int i = 0; i < frm[node].size(); i++)
	{
		if (!iscirc[frm[node][i]])
		{
			v.push_back(dfs(frm[node][i]).first);
		}
	}
	for (int i = 0; i < v.size(); i++)flag[v[i]] = true;
	pii z;
	int c = 0;
	for (int i = 0;; i++)
	{
		if (!flag[i] && c == 0)c++, z.first = i;
		else if (!flag[i] && c == 1)
		{
			z.second = i;
			break;
		}
	}
	for (int i = 0; i < v.size(); i++)flag[v[i]] = false;
	return z;
}
void yes()
{
	printf("POSSIBLE\n");
	exit(0);
}
void no()
{
	printf("IMPOSSIBLE\n");
	exit(0);
}
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)
	{
		scanf("%d", &to[i]);
		to[i]--;
		frm[to[i]].push_back(i);
	}
	int now = 0;
	vector<int>circ;
	for (;;)
	{
		if (flag[now])
		{
			int z = now;
			for (;;)
			{
				circ.push_back(z);
				//printf("%d\n", z + 1);
				iscirc[z] = true;
				z = to[z];
				if (z == now)break;
			}
			break;
		}
		flag[now] = true;
		now = to[now];
	}
	fill(flag, flag + num, false);
	vector<pii>v;
	for (int i = 0; i < circ.size(); i++)v.push_back(dfs(circ[i]));
	int t = v[v.size() - 1].first;
	for (int i = 0; i < v.size(); i++)
	{
		//printf("%d %d\n", v[i].first, v[i].second);
		if (v[i].first == t)t = v[i].second;
		else t = v[i].first;
	}
	if (t == v[v.size() - 1].first)yes();
	t = v[v.size() - 1].second;
	for (int i = 0; i < v.size(); i++)
	{
		if (v[i].first == t)t = v[i].second;
		else t = v[i].first;
	}
	if (t == v[v.size() - 1].second)yes();
	no();
}
### Prompt

Your challenge is to write 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int to[200000];
vector<int>frm[200000];
bool flag[220000];
bool iscirc[200000];
typedef pair<int, int>pii;
pii dfs(int node)
{
	vector<int>v;
	for (int i = 0; i < frm[node].size(); i++)
	{
		if (!iscirc[frm[node][i]])
		{
			v.push_back(dfs(frm[node][i]).first);
		}
	}
	for (int i = 0; i < v.size(); i++)flag[v[i]] = true;
	pii z;
	int c = 0;
	for (int i = 0;; i++)
	{
		if (!flag[i] && c == 0)c++, z.first = i;
		else if (!flag[i] && c == 1)
		{
			z.second = i;
			break;
		}
	}
	for (int i = 0; i < v.size(); i++)flag[v[i]] = false;
	return z;
}
void yes()
{
	printf("POSSIBLE\n");
	exit(0);
}
void no()
{
	printf("IMPOSSIBLE\n");
	exit(0);
}
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)
	{
		scanf("%d", &to[i]);
		to[i]--;
		frm[to[i]].push_back(i);
	}
	int now = 0;
	vector<int>circ;
	for (;;)
	{
		if (flag[now])
		{
			int z = now;
			for (;;)
			{
				circ.push_back(z);
				//printf("%d\n", z + 1);
				iscirc[z] = true;
				z = to[z];
				if (z == now)break;
			}
			break;
		}
		flag[now] = true;
		now = to[now];
	}
	fill(flag, flag + num, false);
	vector<pii>v;
	for (int i = 0; i < circ.size(); i++)v.push_back(dfs(circ[i]));
	int t = v[v.size() - 1].first;
	for (int i = 0; i < v.size(); i++)
	{
		//printf("%d %d\n", v[i].first, v[i].second);
		if (v[i].first == t)t = v[i].second;
		else t = v[i].first;
	}
	if (t == v[v.size() - 1].first)yes();
	t = v[v.size() - 1].second;
	for (int i = 0; i < v.size(); i++)
	{
		if (v[i].first == t)t = v[i].second;
		else t = v[i].first;
	}
	if (t == v[v.size() - 1].second)yes();
	no();
}
```