#include <bits/stdc++.h>

const int N = 200005;
std::vector<int> v[N];
int n, s, p[N], ind[N], val[N], x[N], y[N];

int mex(std::vector<int> &v) {
	std::sort(v.begin(), v.end()); int t = 0;
	for (; t < v.size() && v[t] == t; ++t);
	return t;
}
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> p[i], ++ind[p[i]];
	std::queue<int> q; for (int i = 1; i <= n; ++i) if (!ind[i]) q.push(i);
	while (q.size()) {
		int x = q.front(); q.pop(), val[x] = mex(v[x]);
		v[p[x]].emplace_back(val[x]);
		if (!--ind[p[x]]) q.push(p[x]); 
	}
	for (int i = 1; i <= n; ++i) if (ind[i])
		s = i, x[i] = mex(v[i]), v[i].emplace_back(x[i]), y[i] = mex(v[i]);
	int val = x[s];
	for (int t = p[s]; ; t = p[t]) {
		val = val == x[t] ? y[t] : x[t];
		if (t == s) break;
	}
	if (val == x[s]) return std::cout << "POSSIBLE\n", 0;
	val = y[s];
	for (int t = p[s]; ; t = p[t]) {
		val = val == x[t] ? y[t] : x[t];
		if (t == s) break;
	}
	if (val == y[s]) return std::cout << "POSSIBLE\n", 0;
	std::cout << "IMPOSSIBLE\n";
	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 <bits/stdc++.h>

const int N = 200005;
std::vector<int> v[N];
int n, s, p[N], ind[N], val[N], x[N], y[N];

int mex(std::vector<int> &v) {
	std::sort(v.begin(), v.end()); int t = 0;
	for (; t < v.size() && v[t] == t; ++t);
	return t;
}
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> p[i], ++ind[p[i]];
	std::queue<int> q; for (int i = 1; i <= n; ++i) if (!ind[i]) q.push(i);
	while (q.size()) {
		int x = q.front(); q.pop(), val[x] = mex(v[x]);
		v[p[x]].emplace_back(val[x]);
		if (!--ind[p[x]]) q.push(p[x]); 
	}
	for (int i = 1; i <= n; ++i) if (ind[i])
		s = i, x[i] = mex(v[i]), v[i].emplace_back(x[i]), y[i] = mex(v[i]);
	int val = x[s];
	for (int t = p[s]; ; t = p[t]) {
		val = val == x[t] ? y[t] : x[t];
		if (t == s) break;
	}
	if (val == x[s]) return std::cout << "POSSIBLE\n", 0;
	val = y[s];
	for (int t = p[s]; ; t = p[t]) {
		val = val == x[t] ? y[t] : x[t];
		if (t == s) break;
	}
	if (val == y[s]) return std::cout << "POSSIBLE\n", 0;
	std::cout << "IMPOSSIBLE\n";
	return 0;
}
```