#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int g;

int func(int p[], int c, bool chk[]) {
  if (chk[p[c]]) {
    return c;
  }

  chk[p[c]] = true;
  return func(p, p[c], chk);
}

int dfs(vector<int> v[], int c, int r, int a[], bool flag) {
  if (v[c].size() == 0) {
    a[c] = 0;
    return 0;
  }

  vector<int> pool;
  bool has = false;
  for (int i = 0; i < v[c].size(); i++) {
    if (v[c][i] == r) {
      has = true;
    } else {
      pool.push_back(dfs(v, v[c][i], r, a, flag));
    }
  }
  sort(pool.begin(), pool.end());
  unique(pool.begin(), pool.end());
  for (int i = 0; i < pool.size(); i++) {
    if (pool[i] != i) {
      if (has && flag) {
        has = false;
        g = i;
        pool.push_back(i);
        sort(pool.begin(), pool.end());
        continue;
      }
      a[c] = i;
      return i;
    }
  }

  int ret = has && flag ? pool.size() + 1 : pool.size();
  if (has && flag)
    g = pool.size();

  a[c] = ret;
  return ret;
}

int main() {
  int N;
  int p[200000];
  vector<int> v[200000];

  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> p[i];
    v[--p[i]].push_back(i);
  }

  bool chk[200000];
  for (int i = 0; i < N; i++) {
    chk[i] = false;
  }
  chk[0] = true;
  int r = func(p, 0, chk);

  int a[200000];
  int a1 = dfs(v, r, r, a, false);
  if (a[p[r]] != a1) {
    cout << "POSSIBLE" << endl;
    return 0;
  }

  int a2 = dfs(v, r, r, a, true);
  if (a[p[r]] != a2 && g == a2) {
    cout << "POSSIBLE" << endl;
    return 0;
  }

  cout << "IMPOSSIBLE" << endl;
  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
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int g;

int func(int p[], int c, bool chk[]) {
  if (chk[p[c]]) {
    return c;
  }

  chk[p[c]] = true;
  return func(p, p[c], chk);
}

int dfs(vector<int> v[], int c, int r, int a[], bool flag) {
  if (v[c].size() == 0) {
    a[c] = 0;
    return 0;
  }

  vector<int> pool;
  bool has = false;
  for (int i = 0; i < v[c].size(); i++) {
    if (v[c][i] == r) {
      has = true;
    } else {
      pool.push_back(dfs(v, v[c][i], r, a, flag));
    }
  }
  sort(pool.begin(), pool.end());
  unique(pool.begin(), pool.end());
  for (int i = 0; i < pool.size(); i++) {
    if (pool[i] != i) {
      if (has && flag) {
        has = false;
        g = i;
        pool.push_back(i);
        sort(pool.begin(), pool.end());
        continue;
      }
      a[c] = i;
      return i;
    }
  }

  int ret = has && flag ? pool.size() + 1 : pool.size();
  if (has && flag)
    g = pool.size();

  a[c] = ret;
  return ret;
}

int main() {
  int N;
  int p[200000];
  vector<int> v[200000];

  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> p[i];
    v[--p[i]].push_back(i);
  }

  bool chk[200000];
  for (int i = 0; i < N; i++) {
    chk[i] = false;
  }
  chk[0] = true;
  int r = func(p, 0, chk);

  int a[200000];
  int a1 = dfs(v, r, r, a, false);
  if (a[p[r]] != a1) {
    cout << "POSSIBLE" << endl;
    return 0;
  }

  int a2 = dfs(v, r, r, a, true);
  if (a[p[r]] != a2 && g == a2) {
    cout << "POSSIBLE" << endl;
    return 0;
  }

  cout << "IMPOSSIBLE" << endl;
  return 0;
}

```