#include <bits/stdc++.h>
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FORR(i,a,b) for (int i=(a);i>=(b);i--)
#define pb push_back

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef set<int> si;
const int inf = 1e9;
const int mod = 1e9+7;

int p[200005];
vi g[200005];
bool r[200005];
int n;
int possible(int i){
  si tmp;
  for(vi::iterator itr=g[i].begin(); itr!= g[i].end(); itr++){
    tmp.insert(possible(*itr));
  }
  int c=0;
  while(true){
    if (tmp.find(c) == tmp.end()){
      break;
    }
    c++;
  }
  return c;
}
pii possible_ring(int i, int back){
  si tmp;
  bool z=true;
  int a=-1,b=-1,c=0,l;
  for(vi::iterator itr=g[i].begin(); itr!= g[i].end(); itr++){
    if(*itr == back)continue;
    tmp.insert(possible(*itr));
  }
  while(true){
    if (tmp.find(c) == tmp.end()){
      if(z){
        a = c;
        z = false;
      }else{
        b = c;
        return pii(a,b);
      }
    }
    c++;
  }
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  FOR(i, 1, n+1){
    ll q;
    cin >> q;
    p[i] = q;
    g[q].pb(i);
  }
  int c = 1,b;
  while(!r[c]){
    r[c] = true;
    b = c;
    c = p[c];
  }
  int cf = c;
  pii f = possible_ring(c, b);
  pii pp = f, np;
  b = c;
  c = p[c];
  while(true){
    int _a, _b;
    np = possible_ring(c, b);
    _a = (pp.first == np.first) ? np.second : np.first;
    _b = (pp.second == np.first) ? np.second : np.first;
    pp = pii(_a, _b);
    if(c == cf)break;
    b = c;
    c = p[c];
  }
  if(f.first == pp.first || f.second == pp.second){
    cout << "POSSIBLE"<<endl;
  }else{
    cout << "IMPOSSIBLE"<<endl;
  }
  return 0;
}
### Prompt

Generate 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>
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FORR(i,a,b) for (int i=(a);i>=(b);i--)
#define pb push_back

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef set<int> si;
const int inf = 1e9;
const int mod = 1e9+7;

int p[200005];
vi g[200005];
bool r[200005];
int n;
int possible(int i){
  si tmp;
  for(vi::iterator itr=g[i].begin(); itr!= g[i].end(); itr++){
    tmp.insert(possible(*itr));
  }
  int c=0;
  while(true){
    if (tmp.find(c) == tmp.end()){
      break;
    }
    c++;
  }
  return c;
}
pii possible_ring(int i, int back){
  si tmp;
  bool z=true;
  int a=-1,b=-1,c=0,l;
  for(vi::iterator itr=g[i].begin(); itr!= g[i].end(); itr++){
    if(*itr == back)continue;
    tmp.insert(possible(*itr));
  }
  while(true){
    if (tmp.find(c) == tmp.end()){
      if(z){
        a = c;
        z = false;
      }else{
        b = c;
        return pii(a,b);
      }
    }
    c++;
  }
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  FOR(i, 1, n+1){
    ll q;
    cin >> q;
    p[i] = q;
    g[q].pb(i);
  }
  int c = 1,b;
  while(!r[c]){
    r[c] = true;
    b = c;
    c = p[c];
  }
  int cf = c;
  pii f = possible_ring(c, b);
  pii pp = f, np;
  b = c;
  c = p[c];
  while(true){
    int _a, _b;
    np = possible_ring(c, b);
    _a = (pp.first == np.first) ? np.second : np.first;
    _b = (pp.second == np.first) ? np.second : np.first;
    pp = pii(_a, _b);
    if(c == cf)break;
    b = c;
    c = p[c];
  }
  if(f.first == pp.first || f.second == pp.second){
    cout << "POSSIBLE"<<endl;
  }else{
    cout << "IMPOSSIBLE"<<endl;
  }
  return 0;
}
```