#include <iostream>
#include <queue>
#include <vector>
#include <cstring>
#include <vector>
#include <cstdlib>
using namespace std;
int N;
vector<int> G[200005];
vector<int> Gt[200005];
bool viz[200005];
int dist[200005];
int gr[200005];
int nxt[200005];
bool U[200005];
int  V[200005];
vector<int> ord;
queue<int> Q;
int mex(int nod)
{
    int mex=0;
    while(U[mex])mex++;
    for(auto it:G[nod])U[V[it]]=0;
    return mex;
}
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
    {
        int x=i,y;
        cin>>y;
        G[y].push_back(x);
        Gt[x].push_back(y);
        gr[y]++;
    }
    for(int i=1;i<=N;i++)
    {
        if(!gr[i])Q.push(i);
    }
    while(!Q.empty())
    {
        int nod=Q.front();Q.pop();
        for(auto it:Gt[nod])
        {
            gr[it]--;
            if(!gr[it])
                Q.push(it);
        }
        ord.push_back(nod);
    }
    for(auto it:ord)
    {
        for(auto it2:G[it])
            if(!gr[it2])
                U[V[it2]]=1;
        V[it]=mex(it);
    }
    int len=0;
    for(int i=1;i<=N;i++)
    {
        if(gr[i])
        {
            len++;
            for(auto it2:G[i])
                if(!gr[it2])
                    U[V[it2]]=1;
            V[i]=mex(i);
        }
    }
    int nr=0;
    for(int i=1;i<=N;i++)
    {
        if(gr[i])
        {
            if(!U[V[i]])nr++;
            U[V[i]]=1;
        }
    }
    if(nr>=2||len%2==0)cout<<"POSSIBLE";
    else cout<<"IMPOSSIBLE";
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <iostream>
#include <queue>
#include <vector>
#include <cstring>
#include <vector>
#include <cstdlib>
using namespace std;
int N;
vector<int> G[200005];
vector<int> Gt[200005];
bool viz[200005];
int dist[200005];
int gr[200005];
int nxt[200005];
bool U[200005];
int  V[200005];
vector<int> ord;
queue<int> Q;
int mex(int nod)
{
    int mex=0;
    while(U[mex])mex++;
    for(auto it:G[nod])U[V[it]]=0;
    return mex;
}
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
    {
        int x=i,y;
        cin>>y;
        G[y].push_back(x);
        Gt[x].push_back(y);
        gr[y]++;
    }
    for(int i=1;i<=N;i++)
    {
        if(!gr[i])Q.push(i);
    }
    while(!Q.empty())
    {
        int nod=Q.front();Q.pop();
        for(auto it:Gt[nod])
        {
            gr[it]--;
            if(!gr[it])
                Q.push(it);
        }
        ord.push_back(nod);
    }
    for(auto it:ord)
    {
        for(auto it2:G[it])
            if(!gr[it2])
                U[V[it2]]=1;
        V[it]=mex(it);
    }
    int len=0;
    for(int i=1;i<=N;i++)
    {
        if(gr[i])
        {
            len++;
            for(auto it2:G[i])
                if(!gr[it2])
                    U[V[it2]]=1;
            V[i]=mex(i);
        }
    }
    int nr=0;
    for(int i=1;i<=N;i++)
    {
        if(gr[i])
        {
            if(!U[V[i]])nr++;
            U[V[i]]=1;
        }
    }
    if(nr>=2||len%2==0)cout<<"POSSIBLE";
    else cout<<"IMPOSSIBLE";
    return 0;
}

```