#include <algorithm>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;
const int nil = -2;
const int loop = -1;

int main() {
  int N;
  while (cin >> N) {
    vector<vector<int>> G(N);
    for (int i = 0; i < N; i++) {
      int p;
      cin >> p;
      --p;
      G[p].push_back(i);
    }
    vector<vector<int>> cand(N);
    vector<int> dp(N, nil);
    function<int(int)> dfs = [&](int v) {
      int &res = dp[v];
      if (res != nil) {
        return res;
      }
      res = loop;
      set<int> S;
      bool hasloop = false;
      for (int u : G[v]) {
        int g = dfs(u);
        if (g == loop) {
          hasloop = true;
        } else {
          S.insert(g);
        }
      }
      int g1 = 0;
      while (S.count(g1)) ++g1;
      int g2 = g1 + 1;
      while (S.count(g2)) ++g2;
      if (!hasloop) {
        return res = g1;
      }
      cand[v].push_back(g1);
      cand[v].push_back(g2);
      return res = loop;
    };
    int r = -1;
    for (int v = 0; v < N; v++) {
      dfs(v);
      if (!cand[v].empty()) {
        r = v;
      }
    }
    bool ans = true;
    if (r != -1) {
      ans = false;
      for (int x : cand[r]) {
        for (int u = 0; u < N; u++) {
          if (!cand[u].empty()) {
            dp[u] = nil;
          }
        }
        dp[r] = x;
        set<int> S;
        for (int u : G[r]) {
          S.insert(dfs(u));
        }
        int y = 0;
        while (S.count(y)) ++y;
        if (y == x) {
          ans = true;
          break;
        }
      }
    }
    cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 <algorithm>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;
const int nil = -2;
const int loop = -1;

int main() {
  int N;
  while (cin >> N) {
    vector<vector<int>> G(N);
    for (int i = 0; i < N; i++) {
      int p;
      cin >> p;
      --p;
      G[p].push_back(i);
    }
    vector<vector<int>> cand(N);
    vector<int> dp(N, nil);
    function<int(int)> dfs = [&](int v) {
      int &res = dp[v];
      if (res != nil) {
        return res;
      }
      res = loop;
      set<int> S;
      bool hasloop = false;
      for (int u : G[v]) {
        int g = dfs(u);
        if (g == loop) {
          hasloop = true;
        } else {
          S.insert(g);
        }
      }
      int g1 = 0;
      while (S.count(g1)) ++g1;
      int g2 = g1 + 1;
      while (S.count(g2)) ++g2;
      if (!hasloop) {
        return res = g1;
      }
      cand[v].push_back(g1);
      cand[v].push_back(g2);
      return res = loop;
    };
    int r = -1;
    for (int v = 0; v < N; v++) {
      dfs(v);
      if (!cand[v].empty()) {
        r = v;
      }
    }
    bool ans = true;
    if (r != -1) {
      ans = false;
      for (int x : cand[r]) {
        for (int u = 0; u < N; u++) {
          if (!cand[u].empty()) {
            dp[u] = nil;
          }
        }
        dp[r] = x;
        set<int> S;
        for (int u : G[r]) {
          S.insert(dfs(u));
        }
        int y = 0;
        while (S.count(y)) ++y;
        if (y == x) {
          ans = true;
          break;
        }
      }
    }
    cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << endl;
  }
  return 0;
}

```