#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x)&(-x)
#define fir first
#define sec second
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define maxd 1000000007
typedef long long ll;
const int N=100000;
const double pi=acos(-1.0);
struct node{
	int to,nxt;
}sq[200200];
int n,fa[200200],mex[200200],all=0,head[200200];
bool vis[200200],cir[200200];

int read()
{
	int x=0,f=1;char ch=getchar();
	while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
	while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
	return x*f;
}

void add(int u,int v)
{
	all++;sq[all].to=v;sq[all].nxt=head[u];head[u]=all;
}

void dfs(int u)
{
	int i;
	for (i=head[u];i;i=sq[i].nxt)
	{
		int v=sq[i].to;
		if (!cir[v]) dfs(v);
	}
	for (i=head[u];i;i=sq[i].nxt)
	{
		int v=sq[i].to;
		if (!cir[v]) vis[mex[v]]=1;
	}
	while (vis[mex[u]]) mex[u]++;
	for (i=head[u];i;i=sq[i].nxt) 
	{
		int v=sq[i].to;
		vis[mex[v]]=0;
	}
}

int main()
{
	n=read();
	rep(i,1,n)
	{
		fa[i]=read();
		add(fa[i],i);
	}
	int u=1,cnt=0;
	while (!vis[u]) {vis[u]=1;u=fa[u];}
	while (!cir[u]) {cir[u]=1;u=fa[u];}
	memset(vis,0,sizeof(vis));
	int maxsg=-1,minsg=n+1;
	rep(i,1,n)
	{
		if (cir[i])
		{
			dfs(i);cnt++;
			maxsg=max(maxsg,mex[i]);minsg=min(minsg,mex[i]); 
		}
	}
	if ((maxsg==minsg) && (cnt&1)) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}

### 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<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x)&(-x)
#define fir first
#define sec second
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define maxd 1000000007
typedef long long ll;
const int N=100000;
const double pi=acos(-1.0);
struct node{
	int to,nxt;
}sq[200200];
int n,fa[200200],mex[200200],all=0,head[200200];
bool vis[200200],cir[200200];

int read()
{
	int x=0,f=1;char ch=getchar();
	while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
	while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
	return x*f;
}

void add(int u,int v)
{
	all++;sq[all].to=v;sq[all].nxt=head[u];head[u]=all;
}

void dfs(int u)
{
	int i;
	for (i=head[u];i;i=sq[i].nxt)
	{
		int v=sq[i].to;
		if (!cir[v]) dfs(v);
	}
	for (i=head[u];i;i=sq[i].nxt)
	{
		int v=sq[i].to;
		if (!cir[v]) vis[mex[v]]=1;
	}
	while (vis[mex[u]]) mex[u]++;
	for (i=head[u];i;i=sq[i].nxt) 
	{
		int v=sq[i].to;
		vis[mex[v]]=0;
	}
}

int main()
{
	n=read();
	rep(i,1,n)
	{
		fa[i]=read();
		add(fa[i],i);
	}
	int u=1,cnt=0;
	while (!vis[u]) {vis[u]=1;u=fa[u];}
	while (!cir[u]) {cir[u]=1;u=fa[u];}
	memset(vis,0,sizeof(vis));
	int maxsg=-1,minsg=n+1;
	rep(i,1,n)
	{
		if (cir[i])
		{
			dfs(i);cnt++;
			maxsg=max(maxsg,mex[i]);minsg=min(minsg,mex[i]); 
		}
	}
	if ((maxsg==minsg) && (cnt&1)) puts("IMPOSSIBLE");
	else puts("POSSIBLE");
	return 0;
}

```