#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
template<typename T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
//end

int n,a[201010],par[201010]; vector<int> g[501010];
bool check(int v,int x){ //cycle
   a[v]=x; int cur=par[v];
   while(v!=cur){
      set<int> st;
      for(int nxt:g[cur])st.insert(a[nxt]);
      a[cur]=0;
      while(st.count(a[cur]))a[cur]++;
      cur=par[cur];
   }
   int t=0; set<int> st;
   for(int nxt:g[v])st.insert(a[nxt]);
   while(st.count(t))t++;
   return t==x;
}

int main(){
   int n; cin>>n;
   rep(i,0,n){
      cin>>par[i]; par[i]--;
      g[par[i]].push_back(i);
   }
   queue<int> que;
   rep(i,0,n){
      if(g[i].size()==0)que.push(i);
      else a[i]=-1;
   }
   while(que.size()){
      int v=que.front(); que.pop();
      set<int> st;
      bool f=0;
      for(int nxt:g[v]){
         if(a[nxt]<0){
            f=1; break;
         }
         st.insert(a[nxt]);
      }
      if(f)continue;
      a[v]=0;
      while(st.count(a[v]))a[v]++;
      que.push(par[v]);
   }
   int v=-1; set<int> st;
   rep(i,0,n)if(a[i]==-1){
      v=i; break;
   }
   for(int nxt:g[v])if(a[nxt]!=-1)st.insert(a[nxt]);
   int x=0;
   while(st.count(x))x++;
   if(check(v,x)){
      puts("POSSIBLE");
      return 0;
   }
   x++;
   while(st.count(x))x++;
   if(check(v,x)){
      puts("POSSIBLE");
      return 0;
   }
   puts("IMPOSSIBLE");
   return 0;
}
### Prompt

In cpp, your task is to solve 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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
template<typename T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
//end

int n,a[201010],par[201010]; vector<int> g[501010];
bool check(int v,int x){ //cycle
   a[v]=x; int cur=par[v];
   while(v!=cur){
      set<int> st;
      for(int nxt:g[cur])st.insert(a[nxt]);
      a[cur]=0;
      while(st.count(a[cur]))a[cur]++;
      cur=par[cur];
   }
   int t=0; set<int> st;
   for(int nxt:g[v])st.insert(a[nxt]);
   while(st.count(t))t++;
   return t==x;
}

int main(){
   int n; cin>>n;
   rep(i,0,n){
      cin>>par[i]; par[i]--;
      g[par[i]].push_back(i);
   }
   queue<int> que;
   rep(i,0,n){
      if(g[i].size()==0)que.push(i);
      else a[i]=-1;
   }
   while(que.size()){
      int v=que.front(); que.pop();
      set<int> st;
      bool f=0;
      for(int nxt:g[v]){
         if(a[nxt]<0){
            f=1; break;
         }
         st.insert(a[nxt]);
      }
      if(f)continue;
      a[v]=0;
      while(st.count(a[v]))a[v]++;
      que.push(par[v]);
   }
   int v=-1; set<int> st;
   rep(i,0,n)if(a[i]==-1){
      v=i; break;
   }
   for(int nxt:g[v])if(a[nxt]!=-1)st.insert(a[nxt]);
   int x=0;
   while(st.count(x))x++;
   if(check(v,x)){
      puts("POSSIBLE");
      return 0;
   }
   x++;
   while(st.count(x))x++;
   if(check(v,x)){
      puts("POSSIBLE");
      return 0;
   }
   puts("IMPOSSIBLE");
   return 0;
}
```