#include<bits/stdc++.h>
using namespace std;

const int N = 2e5 + 9;

vector<int>V[N];
int n, o, h[N], t, la[N], cnt, f[N], f2[N], p;
bool is[N], vis[N], in[N];

void Dfs (int x) {
	vis[x] = is[x] = 1;
	for (auto v : V[x]) {
		if (vis[v]) { if (is[v]) in[h[t = 1] = o = v] = 1; } else Dfs(v);
		if (o) break;
	}
	if (o) {
		if (x == o) o = -1;
		if (o != -1) in[h[++t] = x] = 1;
	}
	is[x] = 0;
}

int jj[N];

void Dfs2 (int x) {
	for (auto v : V[x]) if (!in[v]) Dfs2(v);
	++cnt; for (auto v : V[x]) if (!in[v]) la[f[v]] = cnt;
	while (la[f[x]] == cnt) ++f[x];
	if (in[x]) { f2[x] = f[x] + 1; while (la[f2[x]] == cnt) ++f2[x]; }
}

int main () {
	scanf("%d", &n);
	for (int i = 1, j; i <= n; ++i) scanf("%d", &j), V[j].emplace_back(i);
	for (int i = 1; i <= n && o == 0; ++i) if (!vis[i]) Dfs(i);
	for (int i = 1; i <= t; ++i) Dfs2(h[i]);
	p = f[h[1]];
	for (int i = 2; i <= t; ++i) p = f[h[i]] == p ? f2[h[i]] : f[h[i]];
	p = f[h[1]] == p ? f2[h[1]] : f[h[1]];
	if (p == f[h[1]]) return puts("POSSIBLE"), 0;
	p = f2[h[1]];
	for (int i = 2; i <= t; ++i) p = f[h[i]] == p ? f2[h[i]] : f[h[i]];
	p = f[h[1]] == p ? f2[h[1]] : f[h[1]];
	if (p == f2[h[1]]) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}
### Prompt

Your challenge is to write a cpp solution 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 namespace std;

const int N = 2e5 + 9;

vector<int>V[N];
int n, o, h[N], t, la[N], cnt, f[N], f2[N], p;
bool is[N], vis[N], in[N];

void Dfs (int x) {
	vis[x] = is[x] = 1;
	for (auto v : V[x]) {
		if (vis[v]) { if (is[v]) in[h[t = 1] = o = v] = 1; } else Dfs(v);
		if (o) break;
	}
	if (o) {
		if (x == o) o = -1;
		if (o != -1) in[h[++t] = x] = 1;
	}
	is[x] = 0;
}

int jj[N];

void Dfs2 (int x) {
	for (auto v : V[x]) if (!in[v]) Dfs2(v);
	++cnt; for (auto v : V[x]) if (!in[v]) la[f[v]] = cnt;
	while (la[f[x]] == cnt) ++f[x];
	if (in[x]) { f2[x] = f[x] + 1; while (la[f2[x]] == cnt) ++f2[x]; }
}

int main () {
	scanf("%d", &n);
	for (int i = 1, j; i <= n; ++i) scanf("%d", &j), V[j].emplace_back(i);
	for (int i = 1; i <= n && o == 0; ++i) if (!vis[i]) Dfs(i);
	for (int i = 1; i <= t; ++i) Dfs2(h[i]);
	p = f[h[1]];
	for (int i = 2; i <= t; ++i) p = f[h[i]] == p ? f2[h[i]] : f[h[i]];
	p = f[h[1]] == p ? f2[h[1]] : f[h[1]];
	if (p == f[h[1]]) return puts("POSSIBLE"), 0;
	p = f2[h[1]];
	for (int i = 2; i <= t; ++i) p = f[h[i]] == p ? f2[h[i]] : f[h[i]];
	p = f[h[1]] == p ? f2[h[1]] : f[h[1]];
	if (p == f2[h[1]]) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}
```