#include <bits/stdc++.h>
using namespace std;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
vector<int> ch[200005];
int calc(int x, int p) {
	set<int> S;
	for (int y : ch[x]) {
		if (y == p) continue;
		int v = calc(y, x);
		S.insert(v);
	}
	for (int i = 0;; ++i) {
		if (!S.count(i)) return i;
	}
}
int main() {
	int n;
	scanf("%d", &n);
	vector<int> nex(n);
	for (int i = 0; i < n; ++i) {
		scanf("%d", &nex[i]);
		--nex[i];
		ch[nex[i]].push_back(i);
	}
	vector<bool> used(n);
	int x = 0;
	while (!used[x]) {
		used[x] = true;
		x = nex[x];
	}
	int init = x;
	int len = 0;
	// x is on cycle.
	int mx = INT_MIN;
	int mn = INT_MAX;
	do {
		int v = calc(nex[x], x);
		mx = max(mx, v);
		mn = min(mn, v);
		x = nex[x];
		++len;
	} while (x != init);
	if (mx == mn && len % 2) {
		printf("IMPOSSIBLE\n");
	} else {
		printf("POSSIBLE\n");
	}
}

### 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>
using namespace std;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
vector<int> ch[200005];
int calc(int x, int p) {
	set<int> S;
	for (int y : ch[x]) {
		if (y == p) continue;
		int v = calc(y, x);
		S.insert(v);
	}
	for (int i = 0;; ++i) {
		if (!S.count(i)) return i;
	}
}
int main() {
	int n;
	scanf("%d", &n);
	vector<int> nex(n);
	for (int i = 0; i < n; ++i) {
		scanf("%d", &nex[i]);
		--nex[i];
		ch[nex[i]].push_back(i);
	}
	vector<bool> used(n);
	int x = 0;
	while (!used[x]) {
		used[x] = true;
		x = nex[x];
	}
	int init = x;
	int len = 0;
	// x is on cycle.
	int mx = INT_MIN;
	int mn = INT_MAX;
	do {
		int v = calc(nex[x], x);
		mx = max(mx, v);
		mn = min(mn, v);
		x = nex[x];
		++len;
	} while (x != init);
	if (mx == mn && len % 2) {
		printf("IMPOSSIBLE\n");
	} else {
		printf("POSSIBLE\n");
	}
}

```