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

int main() {
    int n; cin >> n;
    vector<int> par(n), a(n, -1);
    vector<vector<int>> g(n);
    for (int i = 0; i < n; i++) {
        cin >> par[i], par[i]--;
        g[par[i]].emplace_back(i);
    }
    queue<int> que;
    for (int i = 0; i < n; i++) if (g[i].size() == 0) que.emplace(i);   // 葉
    while (!que.empty()) {
        int u = que.front(); que.pop();
        if (a[u] >= 0) continue;
        set<int> s;
        bool on_cycle = false;
        for (int v: g[u]) {
            if (a[v] == -1) on_cycle = true;
            s.emplace(a[v]);
        }
        if (on_cycle) continue;
        int g = 0; while (s.count(g)) g++;
        a[u] = g;
        que.emplace(par[u]);
    }
    auto check = [&](int st, int val) {
        int u = par[st];
        int prev = val;
        while (u != st) {
            set<int> s; s.insert(prev);
            for (int v: g[u]) s.insert(a[v]);
            int g = 0; while (s.count(g)) g++;
            prev = a[u] = g;
            u = par[u];
        }
        set<int> s; s.insert(prev);
        for (int v: g[u]) s.insert(a[v]);
        int g = 0; while (s.count(g)) g++;
        a[st] = g;
        return g == val;
    };
    int u;
    for (int i = 0; i < n; i++) if (a[i] == -1) u = i;
    set<int> s;
    for (int v: g[u]) s.emplace(a[v]);
    int g1 = 0;         while (s.count(g1)) g1++;
    int g2 = g1 + 1;    while (s.count(g2)) g2++;
    if (check(u, g1) || check(u, g2)) cout << "POSSIBLE" << endl;
    else cout << "IMPOSSIBLE" << endl;
    return 0;
}
### Prompt

Please provide a Cpp coded solution 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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <set>
using namespace std;

int main() {
    int n; cin >> n;
    vector<int> par(n), a(n, -1);
    vector<vector<int>> g(n);
    for (int i = 0; i < n; i++) {
        cin >> par[i], par[i]--;
        g[par[i]].emplace_back(i);
    }
    queue<int> que;
    for (int i = 0; i < n; i++) if (g[i].size() == 0) que.emplace(i);   // 葉
    while (!que.empty()) {
        int u = que.front(); que.pop();
        if (a[u] >= 0) continue;
        set<int> s;
        bool on_cycle = false;
        for (int v: g[u]) {
            if (a[v] == -1) on_cycle = true;
            s.emplace(a[v]);
        }
        if (on_cycle) continue;
        int g = 0; while (s.count(g)) g++;
        a[u] = g;
        que.emplace(par[u]);
    }
    auto check = [&](int st, int val) {
        int u = par[st];
        int prev = val;
        while (u != st) {
            set<int> s; s.insert(prev);
            for (int v: g[u]) s.insert(a[v]);
            int g = 0; while (s.count(g)) g++;
            prev = a[u] = g;
            u = par[u];
        }
        set<int> s; s.insert(prev);
        for (int v: g[u]) s.insert(a[v]);
        int g = 0; while (s.count(g)) g++;
        a[st] = g;
        return g == val;
    };
    int u;
    for (int i = 0; i < n; i++) if (a[i] == -1) u = i;
    set<int> s;
    for (int v: g[u]) s.emplace(a[v]);
    int g1 = 0;         while (s.count(g1)) g1++;
    int g2 = g1 + 1;    while (s.count(g2)) g2++;
    if (check(u, g1) || check(u, g2)) cout << "POSSIBLE" << endl;
    else cout << "IMPOSSIBLE" << endl;
    return 0;
}
```