#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPD(i,a,b) for (int i = (a); i >= (b); --i)
#define FORI(i,n) REP(i,1,n)
#define FOR(i,n) REP(i,0,int(n)-1)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define ll long long
#define SZ(x) int((x).size())
#define DBG(v) cerr << #v << " = " << (v) << endl;
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define fi first
#define se second

const int N = 200200;

int n;
int vis[N], val[N], outdeg[N], p[N];
set<int> vs[N];
int la;

int mex(set<int> &s) {
	set<int>::iterator it=s.begin();
	int ret=0;
	while (it != s.end()) {
		if (*it != ret) break;
		it++;
		ret++;
	}
	return ret;
}

void dfs(int u) {
	vis[u] = 1;
	val[u] = mex(vs[u]);
	vs[p[u]].insert(val[u]);
	outdeg[p[u]]--;
	if (outdeg[p[u]]==0) dfs(p[u]);
}

bool ok(int u, int vv) {
	int v = u;
	val[u] = vv;
	while (p[v] != u) {
		set<int> st = vs[p[v]];
		st.insert(val[v]);
		int vc = mex(st);
		val[p[v]] = vc;
		vis[v] = 1;
		v = p[v];
	}
	la = v;
	return val[u] != val[v];
}

int main() {
	scanf("%d", &n);
	FOR(i,n) scanf("%d", &p[i]);
	FOR(i,n) p[i]--;
	FOR(i,n) outdeg[p[i]]++;
	FOR(i,n) if (!vis[i] && outdeg[i]==0) dfs(i);
	FOR(i,n) if (!vis[i]) {
		int v1 = mex(vs[i]);
		if (!ok(i, v1)) {
			vs[i].insert(v1);
			int v2 = mex(vs[i]);
			if (!ok(i, v2) || val[la] != v1) {
				printf("IMPOSSIBLE\n");
				return 0;
			}
		}
	}
	printf("POSSIBLE\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 REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPD(i,a,b) for (int i = (a); i >= (b); --i)
#define FORI(i,n) REP(i,1,n)
#define FOR(i,n) REP(i,0,int(n)-1)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define ll long long
#define SZ(x) int((x).size())
#define DBG(v) cerr << #v << " = " << (v) << endl;
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define fi first
#define se second

const int N = 200200;

int n;
int vis[N], val[N], outdeg[N], p[N];
set<int> vs[N];
int la;

int mex(set<int> &s) {
	set<int>::iterator it=s.begin();
	int ret=0;
	while (it != s.end()) {
		if (*it != ret) break;
		it++;
		ret++;
	}
	return ret;
}

void dfs(int u) {
	vis[u] = 1;
	val[u] = mex(vs[u]);
	vs[p[u]].insert(val[u]);
	outdeg[p[u]]--;
	if (outdeg[p[u]]==0) dfs(p[u]);
}

bool ok(int u, int vv) {
	int v = u;
	val[u] = vv;
	while (p[v] != u) {
		set<int> st = vs[p[v]];
		st.insert(val[v]);
		int vc = mex(st);
		val[p[v]] = vc;
		vis[v] = 1;
		v = p[v];
	}
	la = v;
	return val[u] != val[v];
}

int main() {
	scanf("%d", &n);
	FOR(i,n) scanf("%d", &p[i]);
	FOR(i,n) p[i]--;
	FOR(i,n) outdeg[p[i]]++;
	FOR(i,n) if (!vis[i] && outdeg[i]==0) dfs(i);
	FOR(i,n) if (!vis[i]) {
		int v1 = mex(vs[i]);
		if (!ok(i, v1)) {
			vs[i].insert(v1);
			int v2 = mex(vs[i]);
			if (!ok(i, v2) || val[la] != v1) {
				printf("IMPOSSIBLE\n");
				return 0;
			}
		}
	}
	printf("POSSIBLE\n");
	return 0;
}

```