#include <cstdio>
const int N=200005;
int fst[N],nxt[N],p[N],q[N],deg[N],f[N];
bool u[N];
void addedge(int x,int y){
	nxt[y]=fst[x];fst[x]=y;
}
int calc(int x,bool bo){
	for (int i=fst[x];i;i=nxt[i]) if (!deg[i]) u[f[i]]=1;
	int y=0;
	for (;u[y];y++);
	if (bo) for (y++;u[y];y++);
	for (int i=fst[x];i;i=nxt[i]) u[f[i]]=0;
	return y;
}
bool check(int st,bool bo){
	for (int i=st;!deg[i];i=p[i]) deg[i]=1;
	f[st]=calc(st,bo);--deg[st];
	for (int i=p[st];i!=st;i=p[i]){
		f[i]=calc(i,0);
		deg[i]=0;
	}
	return calc(st,0)==f[st];
}
int main(){
	int n;
	scanf("%d\n",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&p[i]);
		addedge(p[i],i);
		deg[p[i]]++;
	}
	int head=1,tail=0;
	for (int i=1;i<=n;i++)
		if (!deg[i]) q[++tail]=i;
	for (;head<=tail;){
		int x=q[head++];
		if (!--deg[p[x]]) q[++tail]=p[x];
		f[x]=calc(x,0);
	}
	int st=0;
	for (int i=1;i<=n;i++)
		if (deg[i]) st=i;
	if (check(st,0) || check(st,1)) printf("POSSIBLE\n");
		else printf("IMPOSSIBLE\n");
}
### 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 <cstdio>
const int N=200005;
int fst[N],nxt[N],p[N],q[N],deg[N],f[N];
bool u[N];
void addedge(int x,int y){
	nxt[y]=fst[x];fst[x]=y;
}
int calc(int x,bool bo){
	for (int i=fst[x];i;i=nxt[i]) if (!deg[i]) u[f[i]]=1;
	int y=0;
	for (;u[y];y++);
	if (bo) for (y++;u[y];y++);
	for (int i=fst[x];i;i=nxt[i]) u[f[i]]=0;
	return y;
}
bool check(int st,bool bo){
	for (int i=st;!deg[i];i=p[i]) deg[i]=1;
	f[st]=calc(st,bo);--deg[st];
	for (int i=p[st];i!=st;i=p[i]){
		f[i]=calc(i,0);
		deg[i]=0;
	}
	return calc(st,0)==f[st];
}
int main(){
	int n;
	scanf("%d\n",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&p[i]);
		addedge(p[i],i);
		deg[p[i]]++;
	}
	int head=1,tail=0;
	for (int i=1;i<=n;i++)
		if (!deg[i]) q[++tail]=i;
	for (;head<=tail;){
		int x=q[head++];
		if (!--deg[p[x]]) q[++tail]=p[x];
		f[x]=calc(x,0);
	}
	int st=0;
	for (int i=1;i<=n;i++)
		if (deg[i]) st=i;
	if (check(st,0) || check(st,1)) printf("POSSIBLE\n");
		else printf("IMPOSSIBLE\n");
}
```