/*
Cerberus97
Hanit Banga
*/

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false)

typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 2e5 + 5;

int a[N], p[N];
vector <int> g[N];
bool seen[N], cycle[N];

void dfs(int i);
bool solve(int st, int v);

int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		cin >> p[i];
		g[p[i]].pb(i);
	}

	int cur = 1;
	for (int i = 0; i <= n; ++i)
		cur = p[cur];

	cycle[cur] = true;
	int st = cur;
	while (p[cur] != st)
	{
		cur = p[cur];
		cycle[cur] = true;
	}

	for (int i = 1; i <= n; ++i)
		if (!cycle[i] and !seen[i])
			dfs(i);

	int v = 0, w = 0;
	set <int> s;
	for (auto &i : g[st])
		if (!cycle[i])
			s.insert(a[i]);

	while (s.find(v) != s.end())
		++v;

	s.insert(v);
	while (s.find(w) != s.end())
		++w;

	if(solve(st, v) or solve(st, w))
		cout << "POSSIBLE\n";
	else
		cout << "IMPOSSIBLE\n";
}

void dfs(int i)
{
	seen[i] = true;
	set <int> s;
	for (auto &j : g[i])
	{
		if (!seen[j])
			dfs(j);

		s.insert(a[j]);
	}

	a[i] = 0;
	while (s.find(a[i]) != s.end())
		++a[i];
}

bool solve(int st, int v)
{
	a[st] = v;
	int cur = p[st];
	while (true)
	{
		set <int> s;
		for (auto &i : g[cur])
			s.insert(a[i]);

		a[cur] = 0;
		while (s.find(a[cur]) != s.end())
			++a[cur];

		if (cur == st)
			break;

		cur = p[cur];
	}

	return (a[st] == v);
}
### Prompt

Please formulate 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
/*
Cerberus97
Hanit Banga
*/

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false)

typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 2e5 + 5;

int a[N], p[N];
vector <int> g[N];
bool seen[N], cycle[N];

void dfs(int i);
bool solve(int st, int v);

int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		cin >> p[i];
		g[p[i]].pb(i);
	}

	int cur = 1;
	for (int i = 0; i <= n; ++i)
		cur = p[cur];

	cycle[cur] = true;
	int st = cur;
	while (p[cur] != st)
	{
		cur = p[cur];
		cycle[cur] = true;
	}

	for (int i = 1; i <= n; ++i)
		if (!cycle[i] and !seen[i])
			dfs(i);

	int v = 0, w = 0;
	set <int> s;
	for (auto &i : g[st])
		if (!cycle[i])
			s.insert(a[i]);

	while (s.find(v) != s.end())
		++v;

	s.insert(v);
	while (s.find(w) != s.end())
		++w;

	if(solve(st, v) or solve(st, w))
		cout << "POSSIBLE\n";
	else
		cout << "IMPOSSIBLE\n";
}

void dfs(int i)
{
	seen[i] = true;
	set <int> s;
	for (auto &j : g[i])
	{
		if (!seen[j])
			dfs(j);

		s.insert(a[j]);
	}

	a[i] = 0;
	while (s.find(a[i]) != s.end())
		++a[i];
}

bool solve(int st, int v)
{
	a[st] = v;
	int cur = p[st];
	while (true)
	{
		set <int> s;
		for (auto &i : g[cur])
			s.insert(a[i]);

		a[cur] = 0;
		while (s.find(a[cur]) != s.end())
			++a[cur];

		if (cur == st)
			break;

		cur = p[cur];
	}

	return (a[st] == v);
}
```