#include <bits/stdc++.h>
using namespace std;

typedef int weight;
typedef pair<weight, int> P;

class edge {
public:
	int to, cap;
	weight cost;
	int rev;
	edge(int t, int ca, weight co, int r):to(t), cap(ca), cost(co), rev(r){};
};

vector<vector<edge> > G;

void init(int n) {
	G.clear();
	G.resize(n);
}

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

weight min_cost_flow(int s, int t, int f) {
	const int n = G.size();
	weight res = 0;
	vector<int> prevv(n), preve(n);
	vector<weight> h(n, 0);
	while(f > 0) {
		priority_queue<P, vector<P>, greater<P> > que;
		vector<weight> dist(n, INT_MAX);
		dist[s] = 0;
		que.push(P(0, s));
		while(!que.empty()) {
			const P p = que.top();
			que.pop();
			const int v = p.second;
			if(dist[v] < p.first)
				continue;

			for(int i = 0; i < G[v].size(); ++i) {
				const edge &e = G[v][i];
				if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					que.push(P(dist[e.to], e.to));
				}
			}
		}

		if(dist[t] == INT_MAX)
			return -1;

		for(int v = 0; v < n; ++v)
			h[v] += dist[v];

		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 * h[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;
	cin >> n;

	vector<int> p(n);
	for(auto &e : p) cin >> e;

	const int source = n + n;
	const int sink = source + 1;
	init(sink + 1);

	for(int i = 0; i < n; ++i) {
		add_edge(source, i, 1, 0);
		add_edge(i + n, sink, 1, 0);

		for(int j = 0; j < n; ++j) {
			if(p[i] == j + 1) continue;
			add_edge(i, j + n, 1, p[i] * abs(i - j));
		}
	}

	cout << min_cost_flow(source, sink, n) << endl;

	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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;

typedef int weight;
typedef pair<weight, int> P;

class edge {
public:
	int to, cap;
	weight cost;
	int rev;
	edge(int t, int ca, weight co, int r):to(t), cap(ca), cost(co), rev(r){};
};

vector<vector<edge> > G;

void init(int n) {
	G.clear();
	G.resize(n);
}

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

weight min_cost_flow(int s, int t, int f) {
	const int n = G.size();
	weight res = 0;
	vector<int> prevv(n), preve(n);
	vector<weight> h(n, 0);
	while(f > 0) {
		priority_queue<P, vector<P>, greater<P> > que;
		vector<weight> dist(n, INT_MAX);
		dist[s] = 0;
		que.push(P(0, s));
		while(!que.empty()) {
			const P p = que.top();
			que.pop();
			const int v = p.second;
			if(dist[v] < p.first)
				continue;

			for(int i = 0; i < G[v].size(); ++i) {
				const edge &e = G[v][i];
				if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					que.push(P(dist[e.to], e.to));
				}
			}
		}

		if(dist[t] == INT_MAX)
			return -1;

		for(int v = 0; v < n; ++v)
			h[v] += dist[v];

		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 * h[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;
	cin >> n;

	vector<int> p(n);
	for(auto &e : p) cin >> e;

	const int source = n + n;
	const int sink = source + 1;
	init(sink + 1);

	for(int i = 0; i < n; ++i) {
		add_edge(source, i, 1, 0);
		add_edge(i + n, sink, 1, 0);

		for(int j = 0; j < n; ++j) {
			if(p[i] == j + 1) continue;
			add_edge(i, j + n, 1, p[i] * abs(i - j));
		}
	}

	cout << min_cost_flow(source, sink, n) << endl;

	return 0;
}
```