#include <bits/stdc++.h>

using namespace std;

#define MAX_V 310
#define INF 1e9

struct edge{
  int to,cap,cost,rev;
  edge(int to,int cap,int cost,int rev) :
    to(to),cap(cap),cost(cost),rev(rev) {}
};

vector<edge> G[MAX_V];
int dist[MAX_V],prevv[MAX_V],preve[MAX_V];

void add_edge(int from,int to,int cap,int cost){
  G[from].push_back(edge(to,cap,cost,G[to].size()));
  G[to].push_back(edge(from,0,-cost,G[from].size()-1));
}

 
int min_cost_flow(int s,int t,int f){
  int res = 0;
  while(f > 0){
    fill(dist,dist+MAX_V,INF); dist[s] = 0;
    bool update = true;
    while(update){
      update = false;
      for(int v = 0 ; v < MAX_V ; v++){
        if(dist[v] == INF) continue;
        for(int i = 0 ; i < (int)G[v].size() ; i++){
          edge &e = G[v][i];
          if(e.cap > 0 && dist[e.to] > dist[v] + e.cost){
            dist[e.to] = dist[v] + e.cost;
            prevv[e.to] = v; preve[e.to] = i; update = true;
          }
        }
      }
    }
    if(dist[t] == INF) return -1;
    int d = f;
    for(int v = t ; v != s ; v = prevv[v]){
      d = min(d,G[prevv[v]][preve[v]].cap);
    }
    f -= d; res += d * dist[t];
    for(int v = t ; v != s ; v = prevv[v]){
      edge &e = G[prevv[v]][preve[v]];
      e.cap -= d; G[v][e.rev].cap += d;
    }
  }
  return res;
}

int main(){
  int N,p[MAX_V];
  cin >> N;
  for(int i = 0 ; i < N ; i++){
    cin >> p[i];
  }
  int s = 2*N+1,t = s+1;
  for(int i = 0 ; i < N ; i++){
    add_edge(s,i+1,1,0);
    add_edge(N+i+1,t,1,0);
  }
  for(int i = 0 ; i < N ; i++){
    for(int j = 0 ; j < N ; j++){
      if(i+1 == p[j]) continue;
      add_edge(i+1,N+j+1,1,p[j]*abs(i-j));
    }
  }
  cout << min_cost_flow(s,t,N) << endl;
  return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

G --Derangement / Derangement

Story

Person D is doing research on permutations. At one point, D found a permutation class called "Derangement". "It's cool to have a perfect permutation !!! And it starts with D !!!", the person of D who had a crush on Chunibyo decided to rewrite all the permutations given to him.

Problem

Given a permutation of length n, p = (p_1 p_2 ... p_n). The operation of swapping the i-th element and the j-th element requires the cost of (p_i + p_j) \ times | i-j |. At this time, find the minimum cost required to make p into a perfect permutation using only the above swap operation.

However, the fact that the permutation q is a perfect permutation means that q_i \ neq i holds for 1 \ leq i \ leq n. For example, (5 3 1 2 4) is a perfect permutation, but (3 2 5 4 1) is not a perfect permutation because the fourth number is 4.

Input

The input consists of the following format.


n
p_1 p_2 ... p_n

The first row consists of one integer representing the length n of the permutation. The second line consists of n integers, and the i-th integer represents the i-th element of the permutation p. Each is separated by a single blank character.

Constraints:

* 2 \ leq n \ leq 100
* 1 \ leq p_i \ leq n, i \ neq j ⇔ p_i \ neq p_j



Output

Print the minimum cost to sort p into a complete permutation on one line. Be sure to start a new line at the end of the line.

Sample Input 1


Five
1 2 3 5 4

Sample Output 1


7

After swapping 1 and 2, swapping 1 and 3 yields (2 3 1 5 4), which is a complete permutation.

Sample Input 2


Five
5 3 1 2 4

Sample Output 2


0

It is a perfect permutation from the beginning.

Sample Input 3


Ten
1 3 4 5 6 2 7 8 9 10

Sample Output 3


38





Example

Input

5
1 2 3 5 4


Output

7

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

#define MAX_V 310
#define INF 1e9

struct edge{
  int to,cap,cost,rev;
  edge(int to,int cap,int cost,int rev) :
    to(to),cap(cap),cost(cost),rev(rev) {}
};

vector<edge> G[MAX_V];
int dist[MAX_V],prevv[MAX_V],preve[MAX_V];

void add_edge(int from,int to,int cap,int cost){
  G[from].push_back(edge(to,cap,cost,G[to].size()));
  G[to].push_back(edge(from,0,-cost,G[from].size()-1));
}

 
int min_cost_flow(int s,int t,int f){
  int res = 0;
  while(f > 0){
    fill(dist,dist+MAX_V,INF); dist[s] = 0;
    bool update = true;
    while(update){
      update = false;
      for(int v = 0 ; v < MAX_V ; v++){
        if(dist[v] == INF) continue;
        for(int i = 0 ; i < (int)G[v].size() ; i++){
          edge &e = G[v][i];
          if(e.cap > 0 && dist[e.to] > dist[v] + e.cost){
            dist[e.to] = dist[v] + e.cost;
            prevv[e.to] = v; preve[e.to] = i; update = true;
          }
        }
      }
    }
    if(dist[t] == INF) return -1;
    int d = f;
    for(int v = t ; v != s ; v = prevv[v]){
      d = min(d,G[prevv[v]][preve[v]].cap);
    }
    f -= d; res += d * dist[t];
    for(int v = t ; v != s ; v = prevv[v]){
      edge &e = G[prevv[v]][preve[v]];
      e.cap -= d; G[v][e.rev].cap += d;
    }
  }
  return res;
}

int main(){
  int N,p[MAX_V];
  cin >> N;
  for(int i = 0 ; i < N ; i++){
    cin >> p[i];
  }
  int s = 2*N+1,t = s+1;
  for(int i = 0 ; i < N ; i++){
    add_edge(s,i+1,1,0);
    add_edge(N+i+1,t,1,0);
  }
  for(int i = 0 ; i < N ; i++){
    for(int j = 0 ; j < N ; j++){
      if(i+1 == p[j]) continue;
      add_edge(i+1,N+j+1,1,p[j]*abs(i-j));
    }
  }
  cout << min_cost_flow(s,t,N) << endl;
  return 0;
}
```