#include <bits/stdc++.h>
using namespace std;
const int N = 300000;
int nxt[N], cnt[N], val[N], wal[N];
vector <int> bi[N];
queue <int> Q;
int n, rt;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++ i)
    {
        cin >> nxt[i], cnt[nxt[i]] ++;
        bi[nxt[i]].push_back(i);
    }
    for (int i = 1; i <= n; ++ i) if (!cnt[i]) Q.push(i);
    while (!Q.empty())
    {
        int t = Q.front(); Q.pop();
        set <int> S;
        for (int i = 0; i < bi[t].size(); ++ i) S.insert(val[bi[t][i]]);
        for (int i = 0; ; ++ i) if (!S.count(i)) {val[t] = i; break;}
        cnt[nxt[t]] --; if (!cnt[nxt[t]]) Q.push(nxt[t]);
    }
    for (int i = 1; i <= n; ++ i) if (cnt[i])
    {
        set <int> S;
        for (int j = 0; j < bi[i].size(); ++ j) if (!cnt[bi[i][j]]) S.insert(val[bi[i][j]]);
        for (int j = 0; ; ++ j) if (!S.count(j)) {val[i] = j; break;}
        for (int j = val[i] + 1; ; ++ j) if (!S.count(j)) {wal[i] = j; break;}
        if (!rt) rt = i;
    }
    {
        int p = val[rt];
        for (int i = nxt[rt]; i != rt; i = nxt[i])
            if (p == val[i]) p = wal[i];
            else p = val[i];
        if (p != val[rt]) {cout << "POSSIBLE" << endl; return 0;}
    }
    {
        int p = wal[rt];
        for (int i = nxt[rt]; i != rt; i = nxt[i])
            if (p == val[i]) p = wal[i];
            else p = val[i];
        if (p == val[rt]) {cout << "POSSIBLE" << endl; return 0;}
    }
    cout << "IMPOSSIBLE" << endl;
}

### 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 <bits/stdc++.h>
using namespace std;
const int N = 300000;
int nxt[N], cnt[N], val[N], wal[N];
vector <int> bi[N];
queue <int> Q;
int n, rt;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++ i)
    {
        cin >> nxt[i], cnt[nxt[i]] ++;
        bi[nxt[i]].push_back(i);
    }
    for (int i = 1; i <= n; ++ i) if (!cnt[i]) Q.push(i);
    while (!Q.empty())
    {
        int t = Q.front(); Q.pop();
        set <int> S;
        for (int i = 0; i < bi[t].size(); ++ i) S.insert(val[bi[t][i]]);
        for (int i = 0; ; ++ i) if (!S.count(i)) {val[t] = i; break;}
        cnt[nxt[t]] --; if (!cnt[nxt[t]]) Q.push(nxt[t]);
    }
    for (int i = 1; i <= n; ++ i) if (cnt[i])
    {
        set <int> S;
        for (int j = 0; j < bi[i].size(); ++ j) if (!cnt[bi[i][j]]) S.insert(val[bi[i][j]]);
        for (int j = 0; ; ++ j) if (!S.count(j)) {val[i] = j; break;}
        for (int j = val[i] + 1; ; ++ j) if (!S.count(j)) {wal[i] = j; break;}
        if (!rt) rt = i;
    }
    {
        int p = val[rt];
        for (int i = nxt[rt]; i != rt; i = nxt[i])
            if (p == val[i]) p = wal[i];
            else p = val[i];
        if (p != val[rt]) {cout << "POSSIBLE" << endl; return 0;}
    }
    {
        int p = wal[rt];
        for (int i = nxt[rt]; i != rt; i = nxt[i])
            if (p == val[i]) p = wal[i];
            else p = val[i];
        if (p == val[rt]) {cout << "POSSIBLE" << endl; return 0;}
    }
    cout << "IMPOSSIBLE" << endl;
}

```