#include<bits/stdc++.h>
#define maxn 400100
using namespace std;
struct edge{int r,nxt;}e[maxn];
int c[maxn],ptr,a[maxn],vis[maxn],head[maxn],tp,bg,tg[maxn],q[maxn],pre[maxn],esz,f[maxn],n;
void addedge(int u,int v){e[++esz].r=v;e[esz].nxt=head[u];head[u]=esz;}
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
int getmex(){
	sort(c+1,c+ptr+1),ptr=unique(c+1,c+ptr+1)-c-1;
	for(int i=1;i<=ptr;++i)if(c[i]!=i-1)return i-1;
	return ptr;
}
void dfs(int u){
	for(int t=head[u];t;t=e[t].nxt)
		if(!tg[e[t].r])dfs(e[t].r);
	ptr=0;
	for(int t=head[u];t;t=e[t].nxt)
		if(!tg[e[t].r])c[++ptr]=a[e[t].r];
	a[u]=getmex();
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)f[i]=i;
	for(int i=1;i<=n;++i){
		scanf("%d",&pre[i]);
		if(find(i)!=find(pre[i]))
			f[find(i)]=find(pre[i]),addedge(pre[i],i);//,printf("[%d->%d]",i,pre[i]);
		else bg=i,addedge(pre[i],i);
	}
	for(int i=bg;;i=pre[i]){
		q[++tp]=i,tg[i]=true;
		if(pre[i]==bg)break;
	}
	for(int i=1;i<=tp;++i)dfs(q[i]);
	int f1=-1,f2=-1;
	for(int t=head[bg];t;t=e[t].nxt)if(!tg[e[t].r])
		vis[a[e[t].r]]=true;
	for(int i=0;i<=n+10;++i)if(!vis[i]){
		if(f1<0)f1=i;
		else if(f2<0)f2=i;
	}
	a[bg]=f1;
	for(int i=2;i<=tp;++i){
		ptr=0;
		for(int t=head[q[i]];t;t=e[t].nxt)
			c[++ptr]=a[e[t].r];//,printf("<%d,%d>",q[i],e[t].r);
		a[q[i]]=getmex();
	}
//	for(int i=1;i<=tp;++i)printf("{%d,%d}",q[i],a[q[i]]);
	ptr=0;
	for(int t=head[bg];t;t=e[t].nxt)
		c[++ptr]=a[e[t].r];
	if(getmex()==a[bg])return puts("POSSIBLE"),0;
	a[bg]=f2;
	for(int i=2;i<=tp;++i){
		ptr=0;
		for(int t=head[q[i]];t;t=e[t].nxt)
			c[++ptr]=a[e[t].r];
		a[q[i]]=getmex();
	}
	ptr=0;
	for(int t=head[bg];t;t=e[t].nxt)
		c[++ptr]=a[e[t].r];
	if(getmex()==a[bg])return puts("POSSIBLE"),0;
	puts("IMPOSSIBLE");
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
#define maxn 400100
using namespace std;
struct edge{int r,nxt;}e[maxn];
int c[maxn],ptr,a[maxn],vis[maxn],head[maxn],tp,bg,tg[maxn],q[maxn],pre[maxn],esz,f[maxn],n;
void addedge(int u,int v){e[++esz].r=v;e[esz].nxt=head[u];head[u]=esz;}
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
int getmex(){
	sort(c+1,c+ptr+1),ptr=unique(c+1,c+ptr+1)-c-1;
	for(int i=1;i<=ptr;++i)if(c[i]!=i-1)return i-1;
	return ptr;
}
void dfs(int u){
	for(int t=head[u];t;t=e[t].nxt)
		if(!tg[e[t].r])dfs(e[t].r);
	ptr=0;
	for(int t=head[u];t;t=e[t].nxt)
		if(!tg[e[t].r])c[++ptr]=a[e[t].r];
	a[u]=getmex();
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)f[i]=i;
	for(int i=1;i<=n;++i){
		scanf("%d",&pre[i]);
		if(find(i)!=find(pre[i]))
			f[find(i)]=find(pre[i]),addedge(pre[i],i);//,printf("[%d->%d]",i,pre[i]);
		else bg=i,addedge(pre[i],i);
	}
	for(int i=bg;;i=pre[i]){
		q[++tp]=i,tg[i]=true;
		if(pre[i]==bg)break;
	}
	for(int i=1;i<=tp;++i)dfs(q[i]);
	int f1=-1,f2=-1;
	for(int t=head[bg];t;t=e[t].nxt)if(!tg[e[t].r])
		vis[a[e[t].r]]=true;
	for(int i=0;i<=n+10;++i)if(!vis[i]){
		if(f1<0)f1=i;
		else if(f2<0)f2=i;
	}
	a[bg]=f1;
	for(int i=2;i<=tp;++i){
		ptr=0;
		for(int t=head[q[i]];t;t=e[t].nxt)
			c[++ptr]=a[e[t].r];//,printf("<%d,%d>",q[i],e[t].r);
		a[q[i]]=getmex();
	}
//	for(int i=1;i<=tp;++i)printf("{%d,%d}",q[i],a[q[i]]);
	ptr=0;
	for(int t=head[bg];t;t=e[t].nxt)
		c[++ptr]=a[e[t].r];
	if(getmex()==a[bg])return puts("POSSIBLE"),0;
	a[bg]=f2;
	for(int i=2;i<=tp;++i){
		ptr=0;
		for(int t=head[q[i]];t;t=e[t].nxt)
			c[++ptr]=a[e[t].r];
		a[q[i]]=getmex();
	}
	ptr=0;
	for(int t=head[bg];t;t=e[t].nxt)
		c[++ptr]=a[e[t].r];
	if(getmex()==a[bg])return puts("POSSIBLE"),0;
	puts("IMPOSSIBLE");
}
```