#include <bits/stdc++.h>

const int MAXN = 200010;
typedef std::vector<int> VI;
VI out[MAXN];
int A[MAXN], n, fa[MAXN], ins[MAXN], bak;
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int dfs(int u) {
	std::set<int> S;
	for (auto i : out[u]) if (!ins[i]) S.insert(dfs(i));
	int now = 0; while (S.count(now)) ++now;
	return now;
}
bool f[MAXN][2];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> A[i], out[A[i]].push_back(fa[i] = i);
	for (int i = 1; i <= n; ++i) if (find(i) == find(A[i])) {
		memset(fa, 0, n + 1 << 2);
		int now = i;
		do ins[fa[++bak] = now] = true, now = A[now]; while (now != i);
		break;
	} else fa[find(i)] = find(A[i]);
	for (int i = 1; i <= bak; ++i) A[i] = dfs(fa[i]);
	int ans = false; n = bak;
	for (int T = 0; T != 2; ++T) {
		memset(f, 0, sizeof f);
		f[1][T] = true;
		for (int j = 2; j <= n; ++j) {
			for (int k = 0; k < 2; ++k) {
				f[j][0] |= f[j - 1][k] && A[j - 1] + k != A[j];
				f[j][1] |= f[j - 1][k] && A[j - 1] + k == A[j];
			}
		}
		for (int k = 0; k < 2; ++k) {
			if (T == 0) ans |= f[n][k] && A[n] + k != A[1];
			if (T == 1) ans |= f[n][k] && A[n] + k == A[1];
		}
	}
	std::cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << '\n';
	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>

const int MAXN = 200010;
typedef std::vector<int> VI;
VI out[MAXN];
int A[MAXN], n, fa[MAXN], ins[MAXN], bak;
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int dfs(int u) {
	std::set<int> S;
	for (auto i : out[u]) if (!ins[i]) S.insert(dfs(i));
	int now = 0; while (S.count(now)) ++now;
	return now;
}
bool f[MAXN][2];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n;
	for (int i = 1; i <= n; ++i) std::cin >> A[i], out[A[i]].push_back(fa[i] = i);
	for (int i = 1; i <= n; ++i) if (find(i) == find(A[i])) {
		memset(fa, 0, n + 1 << 2);
		int now = i;
		do ins[fa[++bak] = now] = true, now = A[now]; while (now != i);
		break;
	} else fa[find(i)] = find(A[i]);
	for (int i = 1; i <= bak; ++i) A[i] = dfs(fa[i]);
	int ans = false; n = bak;
	for (int T = 0; T != 2; ++T) {
		memset(f, 0, sizeof f);
		f[1][T] = true;
		for (int j = 2; j <= n; ++j) {
			for (int k = 0; k < 2; ++k) {
				f[j][0] |= f[j - 1][k] && A[j - 1] + k != A[j];
				f[j][1] |= f[j - 1][k] && A[j - 1] + k == A[j];
			}
		}
		for (int k = 0; k < 2; ++k) {
			if (T == 0) ans |= f[n][k] && A[n] + k != A[1];
			if (T == 1) ans |= f[n][k] && A[n] + k == A[1];
		}
	}
	std::cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << '\n';
	return 0;
}

```