#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 200020
using namespace std;
int read(){
	int nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
int n,m,fs[M],nt[M],to[M],fa[M],tmp,cnt,Mx,Mn,v[M];
bool vis[M],lp[M],hv[M];
void link(int x,int y){nt[tmp]=fs[x],fs[x]=tmp,to[tmp++]=y;}
void dfs(int x){
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) dfs(to[i]);
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) hv[v[to[i]]]=true;
	while(hv[v[x]]) v[x]++;
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) hv[v[to[i]]]=false;
}
int main(){
	n=read(),memset(fs,-1,sizeof(fs)),Mn=n+1;
	for(int i=1;i<=n;i++) fa[i]=read(),link(fa[i],i);
	for(m=1;!vis[m];m=fa[m]) vis[m]=true;
	for(;!lp[m];m=fa[m]) cnt++,lp[m]=true;
	for(m=1;m<=n;m++) if(lp[m]) dfs(m),Mx=max(Mx,v[m]),Mn=min(Mn,v[m]);
	puts((cnt&1)&&Mx==Mn?"IMPOSSIBLE":"POSSIBLE"); return 0;
}
### Prompt

Create a solution in cpp for 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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 200020
using namespace std;
int read(){
	int nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
int n,m,fs[M],nt[M],to[M],fa[M],tmp,cnt,Mx,Mn,v[M];
bool vis[M],lp[M],hv[M];
void link(int x,int y){nt[tmp]=fs[x],fs[x]=tmp,to[tmp++]=y;}
void dfs(int x){
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) dfs(to[i]);
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) hv[v[to[i]]]=true;
	while(hv[v[x]]) v[x]++;
	for(int i=fs[x];i!=-1;i=nt[i]) if(!lp[to[i]]) hv[v[to[i]]]=false;
}
int main(){
	n=read(),memset(fs,-1,sizeof(fs)),Mn=n+1;
	for(int i=1;i<=n;i++) fa[i]=read(),link(fa[i],i);
	for(m=1;!vis[m];m=fa[m]) vis[m]=true;
	for(;!lp[m];m=fa[m]) cnt++,lp[m]=true;
	for(m=1;m<=n;m++) if(lp[m]) dfs(m),Mx=max(Mx,v[m]),Mn=min(Mn,v[m]);
	puts((cnt&1)&&Mx==Mn?"IMPOSSIBLE":"POSSIBLE"); return 0;
}
```