#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int max_n=200005;
int vis[max_n],tot;
vector<int> stk;
int n;
vector<int> G[max_n];
vector<int> cycle;
void dfs(int u){
    stk.push_back(u);
    vis[u]=tot;
    for(auto v:G[u]){
        if(!vis[v])dfs(v);
        else if(vis[v]==tot){
            while(1){
                int x=stk.back();stk.pop_back();
                cycle.push_back(x);
                if(x==v)break;
            }
            return;
        }
    }
    if(!stk.empty())stk.pop_back();
}
int sg[max_n];
int SG(int u){
    if(~sg[u])return sg[u];
    set<int> s;
    for(auto v:G[u])if(!vis[v])s.insert(SG(v));
    for(int i=0;;i++)if(s.count(i)==0)return sg[u]=i;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        int x;scanf("%d",&x);
        G[x].push_back(i);
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            ++tot;
            dfs(i);
        }
    }
    reverse(cycle.begin(),cycle.end());
    if(cycle.empty())printf("POSSIBLE\n");
    else{
        for(int i=1;i<=n;i++)vis[i]=0;
        for(int i=1;i<=n;i++)sg[i]=-1;
        for(auto x:cycle)vis[x]=1;
        for(auto &x:cycle)x=SG(x);
        bool f=false;
        for(int t=0;t<2;t++){
            int pre=cycle.back()+t;
            for(auto x:cycle){
                if(pre==x)pre=x+1;
                else pre=x;
            }
            f|=pre==cycle.back()+t;
        }
        if(f)printf("POSSIBLE\n");
        else printf("IMPOSSIBLE\n");
    }
    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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int max_n=200005;
int vis[max_n],tot;
vector<int> stk;
int n;
vector<int> G[max_n];
vector<int> cycle;
void dfs(int u){
    stk.push_back(u);
    vis[u]=tot;
    for(auto v:G[u]){
        if(!vis[v])dfs(v);
        else if(vis[v]==tot){
            while(1){
                int x=stk.back();stk.pop_back();
                cycle.push_back(x);
                if(x==v)break;
            }
            return;
        }
    }
    if(!stk.empty())stk.pop_back();
}
int sg[max_n];
int SG(int u){
    if(~sg[u])return sg[u];
    set<int> s;
    for(auto v:G[u])if(!vis[v])s.insert(SG(v));
    for(int i=0;;i++)if(s.count(i)==0)return sg[u]=i;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        int x;scanf("%d",&x);
        G[x].push_back(i);
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            ++tot;
            dfs(i);
        }
    }
    reverse(cycle.begin(),cycle.end());
    if(cycle.empty())printf("POSSIBLE\n");
    else{
        for(int i=1;i<=n;i++)vis[i]=0;
        for(int i=1;i<=n;i++)sg[i]=-1;
        for(auto x:cycle)vis[x]=1;
        for(auto &x:cycle)x=SG(x);
        bool f=false;
        for(int t=0;t<2;t++){
            int pre=cycle.back()+t;
            for(auto x:cycle){
                if(pre==x)pre=x+1;
                else pre=x;
            }
            f|=pre==cycle.back()+t;
        }
        if(f)printf("POSSIBLE\n");
        else printf("IMPOSSIBLE\n");
    }
    return 0;
}
```