#include <cstdio>
#include <cstring>
#include <algorithm>
#define last last2

using namespace std;

struct Edge {
  int t,next;
  Edge() {}
  Edge(int a,int b):t(a),next(b) {}
};

Edge e[200005];
int head[200005];

int now[200005];
bool cir[200005],vis[200005];

int f[200005],id[200005];

void dfs(int x,int fa) {
  for(int i=head[x];i;i=e[i].next)
    if (e[i].t!=fa) dfs(e[i].t,x);
  for(int i=head[x];i;i=e[i].next) id[f[e[i].t]]=x;
  while (id[f[x]]==x) f[x]++;
}

int a[200005],b[200005];

bool calc(int rt,int x) {
  int last=x,t=rt;
  do {
  	t=now[t];
    last=(a[t]==last)?b[t]:a[t];
  } while (t!=rt);
  return last==x;
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) {
  	int x;
  	scanf("%d",&x);
  	now[i]=x;
  	e[i]=Edge(i,head[x]);
  	head[x]=i;
  }
  int rt=1;
  do {
  	vis[rt]=1;
  	rt=now[rt];
  } while (!vis[rt]);
  cir[rt]=1;
  for(int i=now[rt];i!=rt;i=now[i]) cir[i]=1;
  for(int i=1;i<=n;i++)
    if (cir[i]) {
    	for(int j=head[i];j;j=e[j].next)
    	  if (!cir[e[j].t]) dfs(e[j].t,i);
		for(int j=head[i];j;j=e[j].next)
		  if (!cir[e[j].t]) id[f[e[j].t]]=i;
		while (id[a[i]]==i) a[i]++;
		b[i]=a[i]+1;
		while (id[b[i]]==i) b[i]++;
	}
  puts((calc(rt,a[rt])|calc(rt,b[rt]))?"POSSIBLE":"IMPOSSIBLE");
  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 <cstdio>
#include <cstring>
#include <algorithm>
#define last last2

using namespace std;

struct Edge {
  int t,next;
  Edge() {}
  Edge(int a,int b):t(a),next(b) {}
};

Edge e[200005];
int head[200005];

int now[200005];
bool cir[200005],vis[200005];

int f[200005],id[200005];

void dfs(int x,int fa) {
  for(int i=head[x];i;i=e[i].next)
    if (e[i].t!=fa) dfs(e[i].t,x);
  for(int i=head[x];i;i=e[i].next) id[f[e[i].t]]=x;
  while (id[f[x]]==x) f[x]++;
}

int a[200005],b[200005];

bool calc(int rt,int x) {
  int last=x,t=rt;
  do {
  	t=now[t];
    last=(a[t]==last)?b[t]:a[t];
  } while (t!=rt);
  return last==x;
}

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) {
  	int x;
  	scanf("%d",&x);
  	now[i]=x;
  	e[i]=Edge(i,head[x]);
  	head[x]=i;
  }
  int rt=1;
  do {
  	vis[rt]=1;
  	rt=now[rt];
  } while (!vis[rt]);
  cir[rt]=1;
  for(int i=now[rt];i!=rt;i=now[i]) cir[i]=1;
  for(int i=1;i<=n;i++)
    if (cir[i]) {
    	for(int j=head[i];j;j=e[j].next)
    	  if (!cir[e[j].t]) dfs(e[j].t,i);
		for(int j=head[i];j;j=e[j].next)
		  if (!cir[e[j].t]) id[f[e[j].t]]=i;
		while (id[a[i]]==i) a[i]++;
		b[i]=a[i]+1;
		while (id[b[i]]==i) b[i]++;
	}
  puts((calc(rt,a[rt])|calc(rt,b[rt]))?"POSSIBLE":"IMPOSSIBLE");
  return 0;
}
```