#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)

const char* possible = "POSSIBLE";
const char* impossible = "IMPOSSIBLE";

int n;
int p[252521];
vector<int> children[252521];

int cycle[252521];
int m = 0;
bool cycled[252521];

int calcorder[252521];
int calcorderhead = 0;
int calcordertail = 0;

int it = 0;

#define MAX_DEG 19
int val[252521];
bool deg[MAX_DEG+2];

void calc(int x){
    REP(i,children[x].size()){
        int v = val[children[x][i]];
        deg[v]=true;
    }
    REP(d,MAX_DEG+1)if(!deg[d]){
        val[x] = d;
        break;
    }
    REP(i,MAX_DEG+2)deg[i]=false;
}

int main(void){
    scanf("%d",&n);
    REP(i,n){
        scanf("%d",p+i);
        p[i]--;
        children[p[i]].push_back(i);
    }
    REP(i,n)it = p[it];
    do{
        cycle[m++] = it;
        cycled[it] = true;
        it = p[it];
    }while(cycle[0]!=it);
    REP(i,m){
        int x = cycle[i];
        REP(j,children[x].size()){
            int y = children[x][j];
            if(cycled[y])continue;
            calcorder[calcordertail++] = y;
        }
    }
    while(calcorderhead != calcordertail){
        int x = calcorder[calcorderhead++];
        REP(j,children[x].size()){
            calcorder[calcordertail++] = children[x][j];
        }
    }
    REP(i,n)val[i] = MAX_DEG+1;
    while(calcordertail--){
        int x = calcorder[calcordertail];
        calc(x);
    }
    // first challenge
    REP(i,m)calc(cycle[i]);
    int tst = val[cycle[0]];
    calc(cycle[0]);
    if(tst == val[cycle[0]]){
        puts(possible);
        return 0;
    }
    REP(i,m-1)calc(cycle[i+1]);
    tst = val[cycle[0]];
    calc(cycle[0]);
    if(tst == val[cycle[0]]){
        puts(possible);
    }else{
        puts(impossible);
    }
    return 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>
using namespace std;
#define REP(i,n) for(int i=0;i<(n);i++)

const char* possible = "POSSIBLE";
const char* impossible = "IMPOSSIBLE";

int n;
int p[252521];
vector<int> children[252521];

int cycle[252521];
int m = 0;
bool cycled[252521];

int calcorder[252521];
int calcorderhead = 0;
int calcordertail = 0;

int it = 0;

#define MAX_DEG 19
int val[252521];
bool deg[MAX_DEG+2];

void calc(int x){
    REP(i,children[x].size()){
        int v = val[children[x][i]];
        deg[v]=true;
    }
    REP(d,MAX_DEG+1)if(!deg[d]){
        val[x] = d;
        break;
    }
    REP(i,MAX_DEG+2)deg[i]=false;
}

int main(void){
    scanf("%d",&n);
    REP(i,n){
        scanf("%d",p+i);
        p[i]--;
        children[p[i]].push_back(i);
    }
    REP(i,n)it = p[it];
    do{
        cycle[m++] = it;
        cycled[it] = true;
        it = p[it];
    }while(cycle[0]!=it);
    REP(i,m){
        int x = cycle[i];
        REP(j,children[x].size()){
            int y = children[x][j];
            if(cycled[y])continue;
            calcorder[calcordertail++] = y;
        }
    }
    while(calcorderhead != calcordertail){
        int x = calcorder[calcorderhead++];
        REP(j,children[x].size()){
            calcorder[calcordertail++] = children[x][j];
        }
    }
    REP(i,n)val[i] = MAX_DEG+1;
    while(calcordertail--){
        int x = calcorder[calcordertail];
        calc(x);
    }
    // first challenge
    REP(i,m)calc(cycle[i]);
    int tst = val[cycle[0]];
    calc(cycle[0]);
    if(tst == val[cycle[0]]){
        puts(possible);
        return 0;
    }
    REP(i,m-1)calc(cycle[i+1]);
    tst = val[cycle[0]];
    calc(cycle[0]);
    if(tst == val[cycle[0]]){
        puts(possible);
    }else{
        puts(impossible);
    }
    return 0;
}

```