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

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 200000;

int n;
int fa[MaxN + 1];

vector<int> adj[MaxN + 1];

int cir_l = 0;
int cir[MaxN + 1];

bool vis[MaxN + 1];

int f[MaxN + 1];

int li_n;
int li[MaxN + 1];

void dfs(const int &u)
{
	for (int v : adj[u])
		if (!vis[v])
			dfs(v);

	li_n = 0;
	li[li_n++] = -1;
	li[li_n++] = MaxN << 8;
	for (int v : adj[u])
		if (!vis[v])
			li[li_n++] = f[v];

	sort(li, li + li_n);
	li_n = unique(li, li + li_n) - li;
	for (int i = 0; i < li_n; ++i)
		if (li[i + 1] != li[i] + 1)
		{
			f[u] = li[i] + 1;
			break;
		}
}

inline bool check()
{
	for (int i = 1; i < cir_l; ++i)
	{
		int u = cir[i];
		int v = cir[i - 1];
		if (f[u] != f[v])
			return true;
	}

	return ~cir_l & 1;
}

int main()
{
	cin >> n;
	for (int v = 1; v <= n; ++v)
	{
		fa[v] = getint();
		adj[fa[v]].push_back(v);
	}

	int sv = 1;
	for (int i = 0; i < n; ++i)
		sv = fa[sv];

	while (!vis[sv])
	{
		cir[cir_l++] = sv;
		vis[sv] = true;
		sv = fa[sv];
	}

	for (int i = 0; i < cir_l; ++i)
		dfs(cir[i]);

	puts(check() ? "POSSIBLE" : "IMPOSSIBLE");

	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 200000;

int n;
int fa[MaxN + 1];

vector<int> adj[MaxN + 1];

int cir_l = 0;
int cir[MaxN + 1];

bool vis[MaxN + 1];

int f[MaxN + 1];

int li_n;
int li[MaxN + 1];

void dfs(const int &u)
{
	for (int v : adj[u])
		if (!vis[v])
			dfs(v);

	li_n = 0;
	li[li_n++] = -1;
	li[li_n++] = MaxN << 8;
	for (int v : adj[u])
		if (!vis[v])
			li[li_n++] = f[v];

	sort(li, li + li_n);
	li_n = unique(li, li + li_n) - li;
	for (int i = 0; i < li_n; ++i)
		if (li[i + 1] != li[i] + 1)
		{
			f[u] = li[i] + 1;
			break;
		}
}

inline bool check()
{
	for (int i = 1; i < cir_l; ++i)
	{
		int u = cir[i];
		int v = cir[i - 1];
		if (f[u] != f[v])
			return true;
	}

	return ~cir_l & 1;
}

int main()
{
	cin >> n;
	for (int v = 1; v <= n; ++v)
	{
		fa[v] = getint();
		adj[fa[v]].push_back(v);
	}

	int sv = 1;
	for (int i = 0; i < n; ++i)
		sv = fa[sv];

	while (!vis[sv])
	{
		cir[cir_l++] = sv;
		vis[sv] = true;
		sv = fa[sv];
	}

	for (int i = 0; i < cir_l; ++i)
		dfs(cir[i]);

	puts(check() ? "POSSIBLE" : "IMPOSSIBLE");

	return 0;
}
```