#include <bits/stdc++.h>
#define MAX_N 200000
using namespace std;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, a[MAX_N+5], b[MAX_N+5];
int into[MAX_N+5], outo[MAX_N+5];
int s, t, f[MAX_N+5], pre[MAX_N+5];
queue <int> que; bool mrk[MAX_N+5];
vector <int> G[MAX_N+5], R[MAX_N+5];
bool chk() {
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (f[u] == f[v]) return false; if (u == t) break;}
	return true;
}
int main() {
	read(n);
	for (int i = 1, fa; i <= n; i++)
		read(fa), 
		G[fa].push_back(i), outo[fa]++, 
		R[i].push_back(fa), into[i]++;
	for (int i = 1; i <= n; i++)
		if (!outo[i]) que.push(i);
	while (!que.empty()) {
		int u = que.front(); que.pop();
		memset(mrk, false, sizeof mrk);
		for (int v : G[u]) mrk[f[v]] = true;
		while (mrk[f[u]]) f[u]++;
		for (int v : R[u]) if (!--outo[v]) que.push(v);
	}
	for (int u = 1; u <= n; u++) if (outo[u]) {
		memset(mrk, false, sizeof mrk), s = u;
		for (int v : G[u]) if (!outo[v]) mrk[f[v]] = true; else pre[v] = u;
		while (mrk[a[u]]) a[u]++; b[u] = a[u]+1; while (mrk[b[u]]) b[u]++;
	}
	f[s] = a[s];
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (v == s) {t = u; break;} f[v] = f[u] == a[v] ? b[v] : a[v];}
	if (f[t] != a[s] && chk()) return puts("POSSIBLE"), 0;
	f[s] = b[s], t = 0;
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (v == s) {t = u; break;} f[v] = f[u] == a[v] ? b[v] : a[v];}
	if (f[t] == a[s] && chk()) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define MAX_N 200000
using namespace std;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, a[MAX_N+5], b[MAX_N+5];
int into[MAX_N+5], outo[MAX_N+5];
int s, t, f[MAX_N+5], pre[MAX_N+5];
queue <int> que; bool mrk[MAX_N+5];
vector <int> G[MAX_N+5], R[MAX_N+5];
bool chk() {
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (f[u] == f[v]) return false; if (u == t) break;}
	return true;
}
int main() {
	read(n);
	for (int i = 1, fa; i <= n; i++)
		read(fa), 
		G[fa].push_back(i), outo[fa]++, 
		R[i].push_back(fa), into[i]++;
	for (int i = 1; i <= n; i++)
		if (!outo[i]) que.push(i);
	while (!que.empty()) {
		int u = que.front(); que.pop();
		memset(mrk, false, sizeof mrk);
		for (int v : G[u]) mrk[f[v]] = true;
		while (mrk[f[u]]) f[u]++;
		for (int v : R[u]) if (!--outo[v]) que.push(v);
	}
	for (int u = 1; u <= n; u++) if (outo[u]) {
		memset(mrk, false, sizeof mrk), s = u;
		for (int v : G[u]) if (!outo[v]) mrk[f[v]] = true; else pre[v] = u;
		while (mrk[a[u]]) a[u]++; b[u] = a[u]+1; while (mrk[b[u]]) b[u]++;
	}
	f[s] = a[s];
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (v == s) {t = u; break;} f[v] = f[u] == a[v] ? b[v] : a[v];}
	if (f[t] != a[s] && chk()) return puts("POSSIBLE"), 0;
	f[s] = b[s], t = 0;
	for (int u = s, v = pre[u]; ; u = v, v = pre[u])
		{if (v == s) {t = u; break;} f[v] = f[u] == a[v] ? b[v] : a[v];}
	if (f[t] == a[s] && chk()) return puts("POSSIBLE"), 0;
	return puts("IMPOSSIBLE"), 0;
}

```