#include <cstring>
#include <cstdio>
using namespace std;

const int N = 2e5 + 10;
struct E {int v, nt;} e[N<<1];
int n, top, cnt, tot, h[N], a[N], vis[N], lst[N], par[N], stk[N];

inline void add(int u, int v) {e[++tot] = (E){v, h[u]}, h[u] = tot;}
template <class T> inline void in(T &x) {
	x = 0; int f = 1; char ch = getchar();
	for (; ch<'0' || ch>'9';) {if (ch=='-') f=-1; ch = getchar();}
	for (; ch>='0' && ch<='9';) x = x*10 + ch-'0', ch = getchar();
	x *= f;
}

inline void dfs(int x) {
	for (int v, i = h[x]; i; i = e[i].nt)
		if (!vis[v = e[i].v]) dfs(v);
	++ cnt;
	for (int v, i = h[x]; i; i = e[i].nt)
		if (!vis[v = e[i].v]) lst[a[v]] = cnt;
	for (a[x] = 0; lst[a[x]] == cnt; ++a[x]);
}

inline void work() {
	int x, p; in(n);
	for (int i = 1; i <= n; ++i) in(par[i]), add(par[i], i);
	for (x = 1; !vis[x]; x = par[x]) vis[stk[++top] = x] = 1;
	for (p = 0; stk[p] != x; ++p); stk[top + 1] = stk[p];
	if ((top - p + 1) % 2 == 0) {puts("POSSIBLE"); return;}
	memset(vis, 0, sizeof vis);
	for (int i = p; i <= top; ++i) vis[stk[i]] = 1;
	for (int i = p; i <= top; ++i) dfs(stk[i]);
	for (int i = p; i <= top; ++i) if (a[stk[i]] != a[stk[i + 1]]) {puts("POSSIBLE"); return;}
	puts("IMPOSSIBLE");
}

int main() {work(); return 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 <cstring>
#include <cstdio>
using namespace std;

const int N = 2e5 + 10;
struct E {int v, nt;} e[N<<1];
int n, top, cnt, tot, h[N], a[N], vis[N], lst[N], par[N], stk[N];

inline void add(int u, int v) {e[++tot] = (E){v, h[u]}, h[u] = tot;}
template <class T> inline void in(T &x) {
	x = 0; int f = 1; char ch = getchar();
	for (; ch<'0' || ch>'9';) {if (ch=='-') f=-1; ch = getchar();}
	for (; ch>='0' && ch<='9';) x = x*10 + ch-'0', ch = getchar();
	x *= f;
}

inline void dfs(int x) {
	for (int v, i = h[x]; i; i = e[i].nt)
		if (!vis[v = e[i].v]) dfs(v);
	++ cnt;
	for (int v, i = h[x]; i; i = e[i].nt)
		if (!vis[v = e[i].v]) lst[a[v]] = cnt;
	for (a[x] = 0; lst[a[x]] == cnt; ++a[x]);
}

inline void work() {
	int x, p; in(n);
	for (int i = 1; i <= n; ++i) in(par[i]), add(par[i], i);
	for (x = 1; !vis[x]; x = par[x]) vis[stk[++top] = x] = 1;
	for (p = 0; stk[p] != x; ++p); stk[top + 1] = stk[p];
	if ((top - p + 1) % 2 == 0) {puts("POSSIBLE"); return;}
	memset(vis, 0, sizeof vis);
	for (int i = p; i <= top; ++i) vis[stk[i]] = 1;
	for (int i = p; i <= top; ++i) dfs(stk[i]);
	for (int i = p; i <= top; ++i) if (a[stk[i]] != a[stk[i + 1]]) {puts("POSSIBLE"); return;}
	puts("IMPOSSIBLE");
}

int main() {work(); return 0;}
```