#include <bits/stdc++.h>
using std::cerr;
using std::endl;

const int N = 2e5 + 10;

int n, fa[N], vis[N], cir[N], cnt, mex[N], buc[N];
std::vector<int> G[N];

void dfs(int x) {
  for (int y : G[x])
    if (!cir[y])
      dfs(y);
  for (int y : G[x])
    if (!cir[y])
      buc[mex[y]] = 1;
  while (buc[mex[x]]) ++mex[x];
  for (int y : G[x])
    if (!cir[y])
      buc[mex[y]] = 0;
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", fa + i), G[fa[i]].push_back(i);
  int x = 1;
  while (!vis[x]) vis[x] = 1, x = fa[x];
  while (!cir[x]) cir[x] = 1, x = fa[x], ++cnt;
  int same = 1, val = -1;
  for (int i = 1; i <= n; ++i)
    if (cir[i]) {
      dfs(i);
      if (val == -1) val = mex[i];
      else if (val != mex[i]) {
        same = 0; break;
      }
    }
  if (same && (cnt & 1)) puts("IMPOSSIBLE");
  else puts("POSSIBLE");
  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 <bits/stdc++.h>
using std::cerr;
using std::endl;

const int N = 2e5 + 10;

int n, fa[N], vis[N], cir[N], cnt, mex[N], buc[N];
std::vector<int> G[N];

void dfs(int x) {
  for (int y : G[x])
    if (!cir[y])
      dfs(y);
  for (int y : G[x])
    if (!cir[y])
      buc[mex[y]] = 1;
  while (buc[mex[x]]) ++mex[x];
  for (int y : G[x])
    if (!cir[y])
      buc[mex[y]] = 0;
}

int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    scanf("%d", fa + i), G[fa[i]].push_back(i);
  int x = 1;
  while (!vis[x]) vis[x] = 1, x = fa[x];
  while (!cir[x]) cir[x] = 1, x = fa[x], ++cnt;
  int same = 1, val = -1;
  for (int i = 1; i <= n; ++i)
    if (cir[i]) {
      dfs(i);
      if (val == -1) val = mex[i];
      else if (val != mex[i]) {
        same = 0; break;
      }
    }
  if (same && (cnt & 1)) puts("IMPOSSIBLE");
  else puts("POSSIBLE");
  return 0;
}

```