#include <bits/stdc++.h>
const int N=200050;
using namespace std;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int t[N],f[N],p[N],st[N],vis[N],cnt,tot,nxt[N],head[N],in[N];
int F(int x){
  ++cnt;
  for (int i=head[x]; i; i=nxt[i])
    t[f[i]]=cnt;
  for (int i=0; ; ++i)
    if (t[i]!=cnt)
      return i;
}
void dfs(int x){
  for (int i=head[x]; i; i=nxt[i])
    dfs(i);
  f[x]=F(x);
}
int main(){
  int n=gi();
  for (int i=1; i<=n; ++i)
    p[i]=gi();
  int k;
  for (k=1; !vis[k]; vis[k]=1,k=p[k]);
  in[st[++tot]=k]=1;
  for (int i=p[k]; i!=k; i=p[i])
    in[st[++tot]=i]=1;
  for (int i=1; i<=n; ++i)
    if (!in[i])
      nxt[i]=head[p[i]],head[p[i]]=i;
  for (int i=1; i<=tot; ++i)
    dfs(st[i]);
  int a=f[st[tot]],b=a+1;
  while(t[b]==cnt) ++b;
  for (int i=1; i<=tot; ++i)
    nxt[st[i]]=head[p[st[i]]],head[p[st[i]]]=st[i];
  f[st[tot]]=a;
  for (int i=1; i<tot; ++i) f[st[i]]=F(st[i]);
  if (f[st[tot-1]]!=a) return puts("POSSIBLE"),0;
  f[st[tot]]=b;
  for (int i=1; i<tot; ++i) f[st[i]]=F(st[i]);
  if (f[st[tot-1]]==a) return puts("POSSIBLE"),0;
  puts("IMPOSSIBLE");
  return 0;
}

### Prompt

Your task is to create 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 <bits/stdc++.h>
const int N=200050;
using namespace std;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int t[N],f[N],p[N],st[N],vis[N],cnt,tot,nxt[N],head[N],in[N];
int F(int x){
  ++cnt;
  for (int i=head[x]; i; i=nxt[i])
    t[f[i]]=cnt;
  for (int i=0; ; ++i)
    if (t[i]!=cnt)
      return i;
}
void dfs(int x){
  for (int i=head[x]; i; i=nxt[i])
    dfs(i);
  f[x]=F(x);
}
int main(){
  int n=gi();
  for (int i=1; i<=n; ++i)
    p[i]=gi();
  int k;
  for (k=1; !vis[k]; vis[k]=1,k=p[k]);
  in[st[++tot]=k]=1;
  for (int i=p[k]; i!=k; i=p[i])
    in[st[++tot]=i]=1;
  for (int i=1; i<=n; ++i)
    if (!in[i])
      nxt[i]=head[p[i]],head[p[i]]=i;
  for (int i=1; i<=tot; ++i)
    dfs(st[i]);
  int a=f[st[tot]],b=a+1;
  while(t[b]==cnt) ++b;
  for (int i=1; i<=tot; ++i)
    nxt[st[i]]=head[p[st[i]]],head[p[st[i]]]=st[i];
  f[st[tot]]=a;
  for (int i=1; i<tot; ++i) f[st[i]]=F(st[i]);
  if (f[st[tot-1]]!=a) return puts("POSSIBLE"),0;
  f[st[tot]]=b;
  for (int i=1; i<tot; ++i) f[st[i]]=F(st[i]);
  if (f[st[tot-1]]==a) return puts("POSSIBLE"),0;
  puts("IMPOSSIBLE");
  return 0;
}

```