#include <cstring>
#include <cstdio>
int fa[200005], val[200005], nxt[200005], ban;
int lst[200005], to[200005], pre[200005], tot; 
bool vis[200005];
inline void add_edge(int u, int v)
{
	to[tot] = v; 
	pre[tot] = lst[u];
	lst[u] = tot++; 
}
int getfa(int u)
{
	if (u == fa[u])
		return u; 
	return fa[u] = getfa(fa[u]); 
}
void dfs(int u)
{
	if (u == ban || lst[u] == -1)
		return; 
	for (int i = lst[u]; ~i; i = pre[i])
		dfs(to[i]); 
	for (int i = lst[u]; ~i; i = pre[i])
		vis[val[to[i]]] = true; 
	for (val[u] = 0; vis[val[u]]; val[u]++); 
	for (int i = lst[u]; ~i; i = pre[i])
		vis[val[to[i]]] = false; 
}
int main()
{
	// freopen("ARC079-F.in", "r", stdin); 
	memset(lst, -1, sizeof(lst)); 
	int n, rt; 
	scanf("%d", &n); 
	for (int i = 0; i < n; i++)
		fa[i] = i; 
	for (int i = 0; i < n; i++)
	{
		scanf("%d", nxt + i); 
		nxt[i]--; 
		if (getfa(i) == getfa(nxt[i]))
			rt = i; 
		else
		{
			fa[getfa(i)] = getfa(nxt[i]); 
			add_edge(nxt[i], i); 
		}
	}
	ban = -1;
	dfs(rt); 
	if (val[rt] == val[nxt[rt]])
	{
		ban = nxt[rt]; 
		val[nxt[rt]]++; 
		dfs(rt); 
		if (val[rt] != val[nxt[rt]] - 1)
		{
			puts("IMPOSSIBLE"); 
			return 0; 
		}
	}
	puts("POSSIBLE"); 
	return 0; 
}

### Prompt

Please formulate 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>
int fa[200005], val[200005], nxt[200005], ban;
int lst[200005], to[200005], pre[200005], tot; 
bool vis[200005];
inline void add_edge(int u, int v)
{
	to[tot] = v; 
	pre[tot] = lst[u];
	lst[u] = tot++; 
}
int getfa(int u)
{
	if (u == fa[u])
		return u; 
	return fa[u] = getfa(fa[u]); 
}
void dfs(int u)
{
	if (u == ban || lst[u] == -1)
		return; 
	for (int i = lst[u]; ~i; i = pre[i])
		dfs(to[i]); 
	for (int i = lst[u]; ~i; i = pre[i])
		vis[val[to[i]]] = true; 
	for (val[u] = 0; vis[val[u]]; val[u]++); 
	for (int i = lst[u]; ~i; i = pre[i])
		vis[val[to[i]]] = false; 
}
int main()
{
	// freopen("ARC079-F.in", "r", stdin); 
	memset(lst, -1, sizeof(lst)); 
	int n, rt; 
	scanf("%d", &n); 
	for (int i = 0; i < n; i++)
		fa[i] = i; 
	for (int i = 0; i < n; i++)
	{
		scanf("%d", nxt + i); 
		nxt[i]--; 
		if (getfa(i) == getfa(nxt[i]))
			rt = i; 
		else
		{
			fa[getfa(i)] = getfa(nxt[i]); 
			add_edge(nxt[i], i); 
		}
	}
	ban = -1;
	dfs(rt); 
	if (val[rt] == val[nxt[rt]])
	{
		ban = nxt[rt]; 
		val[nxt[rt]]++; 
		dfs(rt); 
		if (val[rt] != val[nxt[rt]] - 1)
		{
			puts("IMPOSSIBLE"); 
			return 0; 
		}
	}
	puts("POSSIBLE"); 
	return 0; 
}

```