#include <bits/stdc++.h>
#define possible {cout << "POSSIBLE\n"; return 0;}
#define impossible {cout << "IMPOSSIBLE\n"; return 0;}

using namespace std;

const int Nmax = 2e5 + 5;
int n, i, x, go[Nmax], j, val[Nmax];
bool used[Nmax], active[Nmax];
vector<int> v[Nmax];

bool dfs(int node)
{
    if(active[node]) return 1;
    used[node] = 1;
    active[node] = 1;

    for(auto it : v[node])
        if(dfs(it)) return 1;

    active[node] = 0;
    return 0;
}

void grundy(int node)
{
    for(auto it : v[node])
        if(val[it] == -1) grundy(it);

    for(int i = 0; i <= v[node].size(); ++i) used[i] = 0;
    for(auto it : v[node]) used[val[it]] = 1;

    for(int i = 0; val[node] == -1; ++i)
        if(!used[i]) val[node] = i;
}

int main()
{
   // freopen("input", "r", stdin);
  //  freopen("output", "w", stdout);
    cin.sync_with_stdio(false);

    cin >> n;
    for(i=1; i<=n; ++i)
    {
        cin >> x;
        v[x].push_back(i);
        val[i] = -1;
    }

    for(i=1; i<=n; ++i)
        if(!used[i]) dfs(i);

    for(i=1; i<=n; ++i)
        if(active[i])
            for(auto it : v[i])
                if(!active[it]) grundy(it);
                    else go[i] = it;

    for(i=1; i<=n; ++i)
        if(active[i])
        {
            for(j=0; j<v[i].size(); ++j) used[j] = 0;
            for(auto it : v[i])
                if(!active[it]) used[val[it]] = 1;

            for(j=0; ; ++j)
                if(!used[j]) break;
            val[i] = j;

            grundy(go[i]);
            if(val[go[i]] != val[i]) possible

            for(j=go[i]; j!=i; j = go[j]) val[j] = -1;

            ++val[i];
            grundy(go[i]);
            if(val[go[i]] == val[i] - 1) possible

            impossible
      }

    return 0;
}

### Prompt

Please formulate 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 possible {cout << "POSSIBLE\n"; return 0;}
#define impossible {cout << "IMPOSSIBLE\n"; return 0;}

using namespace std;

const int Nmax = 2e5 + 5;
int n, i, x, go[Nmax], j, val[Nmax];
bool used[Nmax], active[Nmax];
vector<int> v[Nmax];

bool dfs(int node)
{
    if(active[node]) return 1;
    used[node] = 1;
    active[node] = 1;

    for(auto it : v[node])
        if(dfs(it)) return 1;

    active[node] = 0;
    return 0;
}

void grundy(int node)
{
    for(auto it : v[node])
        if(val[it] == -1) grundy(it);

    for(int i = 0; i <= v[node].size(); ++i) used[i] = 0;
    for(auto it : v[node]) used[val[it]] = 1;

    for(int i = 0; val[node] == -1; ++i)
        if(!used[i]) val[node] = i;
}

int main()
{
   // freopen("input", "r", stdin);
  //  freopen("output", "w", stdout);
    cin.sync_with_stdio(false);

    cin >> n;
    for(i=1; i<=n; ++i)
    {
        cin >> x;
        v[x].push_back(i);
        val[i] = -1;
    }

    for(i=1; i<=n; ++i)
        if(!used[i]) dfs(i);

    for(i=1; i<=n; ++i)
        if(active[i])
            for(auto it : v[i])
                if(!active[it]) grundy(it);
                    else go[i] = it;

    for(i=1; i<=n; ++i)
        if(active[i])
        {
            for(j=0; j<v[i].size(); ++j) used[j] = 0;
            for(auto it : v[i])
                if(!active[it]) used[val[it]] = 1;

            for(j=0; ; ++j)
                if(!used[j]) break;
            val[i] = j;

            grundy(go[i]);
            if(val[go[i]] != val[i]) possible

            for(j=go[i]; j!=i; j = go[j]) val[j] = -1;

            ++val[i];
            grundy(go[i]);
            if(val[go[i]] == val[i] - 1) possible

            impossible
      }

    return 0;
}

```