#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=2e5+5;
int p[N],a[N],du[N],q[N];
unordered_set<int>have[N];
int may[N][2];

int get(int x,int j)
{
	while(have[x].count(j))++j;
	return j;
}
int x;
int check(int v)
{
	a[x]=v;
	int y=x;
	while(1)
	{
		int f=p[y];
		if(f==x)break;
		a[f]=may[f][a[y]==may[f][0]];
		y=f;
	}
	return a[y];
}

int main()
{
	int n;
	cin>>n;
	rep(i,1,n){scanf("%d",p+i);++du[p[i]];}
	int tail=0;
	rep(i,1,n)
	if(!du[i])q[++tail]=i;
	rep(head,1,tail)
	{
		int x=q[head];
		a[x]=get(x,0);
		have[p[x]].insert(a[x]);
		if(!--du[p[x]])q[++tail]=p[x];
	}
	x=1;
	while(!du[x])++x;
	for(int y=x;;)
	{
		may[y][0]=get(y,0);
		may[y][1]=get(y,may[y][0]+1);
		if((y=p[y])==x)break;
	}
	if(check(may[x][0])!=may[x][0]||check(may[x][1])==may[x][0])puts("POSSIBLE");
	else puts("IMPOSSIBLE");
}
### Prompt

Develop a solution in CPP 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<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=2e5+5;
int p[N],a[N],du[N],q[N];
unordered_set<int>have[N];
int may[N][2];

int get(int x,int j)
{
	while(have[x].count(j))++j;
	return j;
}
int x;
int check(int v)
{
	a[x]=v;
	int y=x;
	while(1)
	{
		int f=p[y];
		if(f==x)break;
		a[f]=may[f][a[y]==may[f][0]];
		y=f;
	}
	return a[y];
}

int main()
{
	int n;
	cin>>n;
	rep(i,1,n){scanf("%d",p+i);++du[p[i]];}
	int tail=0;
	rep(i,1,n)
	if(!du[i])q[++tail]=i;
	rep(head,1,tail)
	{
		int x=q[head];
		a[x]=get(x,0);
		have[p[x]].insert(a[x]);
		if(!--du[p[x]])q[++tail]=p[x];
	}
	x=1;
	while(!du[x])++x;
	for(int y=x;;)
	{
		may[y][0]=get(y,0);
		may[y][1]=get(y,may[y][0]+1);
		if((y=p[y])==x)break;
	}
	if(check(may[x][0])!=may[x][0]||check(may[x][1])==may[x][0])puts("POSSIBLE");
	else puts("IMPOSSIBLE");
}
```