#include<bits/stdc++.h>
using namespace std;
#define MAXN	200005
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
vector <int> a[MAXN];
int n, f[MAXN], tmp[MAXN], tnp[MAXN];
int nowtype, sg[MAXN], mark[MAXN];
bool visited[MAXN], circle[MAXN], finished[MAXN];
void dfs(int pos) {
	for (unsigned i = 0; i < a[pos].size(); i++)
		dfs(a[pos][i]);
	nowtype++;
	for (unsigned i = 0; i < a[pos].size(); i++)
		mark[sg[a[pos][i]]] = nowtype;
	int ans = 0;
	while (mark[ans] == nowtype) ans++;
	sg[pos] = ans;
}
int main() {
	read(n);
	for (int i = 1; i <= n; i++)
		read(f[i]);
	int pos = 1;
	while (!visited[pos]) {
		visited[pos] = true;
		pos = f[pos];
	}
	while (!circle[pos]) {
		circle[pos] = true;
		pos = f[pos];
	}
	for (int i = 1; i <= n; i++)
		if (!circle[i]) a[f[i]].push_back(i);
	for (int i = 1; i <= n; i++)
		if (circle[i]) dfs(i);
	int last = pos;
	finished[pos] = true;
	tmp[pos] = sg[pos] + 1;
	tnp[pos] = sg[pos];
	pos = f[pos];
	while (!finished[pos]) {
		tmp[pos] = sg[pos] + (sg[pos] == tmp[last]);
		tnp[pos] = sg[pos] + (sg[pos] == tnp[last]);
		finished[pos] = true;
		last = pos; pos = f[pos];
	}
	if (sg[pos] == tmp[last] || tnp[pos] != tnp[last]) printf("POSSIBLE\n");
	else printf("IMPOSSIBLE\n");
	return 0;
}
### Prompt

Develop a solution in CPP 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;
#define MAXN	200005
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
vector <int> a[MAXN];
int n, f[MAXN], tmp[MAXN], tnp[MAXN];
int nowtype, sg[MAXN], mark[MAXN];
bool visited[MAXN], circle[MAXN], finished[MAXN];
void dfs(int pos) {
	for (unsigned i = 0; i < a[pos].size(); i++)
		dfs(a[pos][i]);
	nowtype++;
	for (unsigned i = 0; i < a[pos].size(); i++)
		mark[sg[a[pos][i]]] = nowtype;
	int ans = 0;
	while (mark[ans] == nowtype) ans++;
	sg[pos] = ans;
}
int main() {
	read(n);
	for (int i = 1; i <= n; i++)
		read(f[i]);
	int pos = 1;
	while (!visited[pos]) {
		visited[pos] = true;
		pos = f[pos];
	}
	while (!circle[pos]) {
		circle[pos] = true;
		pos = f[pos];
	}
	for (int i = 1; i <= n; i++)
		if (!circle[i]) a[f[i]].push_back(i);
	for (int i = 1; i <= n; i++)
		if (circle[i]) dfs(i);
	int last = pos;
	finished[pos] = true;
	tmp[pos] = sg[pos] + 1;
	tnp[pos] = sg[pos];
	pos = f[pos];
	while (!finished[pos]) {
		tmp[pos] = sg[pos] + (sg[pos] == tmp[last]);
		tnp[pos] = sg[pos] + (sg[pos] == tnp[last]);
		finished[pos] = true;
		last = pos; pos = f[pos];
	}
	if (sg[pos] == tmp[last] || tnp[pos] != tnp[last]) printf("POSSIBLE\n");
	else printf("IMPOSSIBLE\n");
	return 0;
}
```