p01693 Derangement
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
```python
# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7

class MinCostFlow:
    """ 最小費用流(ダイクストラ版)：O(F*E*logV) """

    INF = 10 ** 18

    def __init__(self, N):
        self.N = N
        self.G = [[] for i in range(N)]

    def add_edge(self, fr, to, cap, cost):
        G = self.G
        G[fr].append([to, cap, cost, len(G[to])])
        G[to].append([fr, 0, -cost, len(G[fr])-1])

    def flow(self, s, t, f):
        from heapq import heappush, heappop

        N = self.N; G = self.G
        INF = MinCostFlow.INF

        res = 0
        H = [0] * N
        prv_v = [0] * N
        prv_e = [0] * N

        while f:
            dist = [INF] * N
            dist[s] = 0
            que = [(0, s)]

            while que:
                c, v = heappop(que)
                if dist[v] < c:
                    continue
                for i, (w, cap, cost, _) in enumerate(G[v]):
                    if cap > 0 and dist[w] > dist[v] + cost + H[v] - H[w]:
                        dist[w] = r = dist[v] + cost + H[v] - H[w]
                        prv_v[w] = v; prv_e[w] = i
                        heappush(que, (r, w))
            if dist[t] == INF:
                return INF

            for i in range(N):
                H[i] += dist[i]

            d = f; v = t
            while v != s:
                d = min(d, G[prv_v[v]][prv_e[v]][1])
                v = prv_v[v]
            f -= d
            res += d * H[t]
            v = t
            while v != s:
                e = G[prv_v[v]][prv_e[v]]
                e[1] -= d
                G[v][e[3]][1] += d
                v = prv_v[v]
        return res

N = INT()
A = LIST()

mcf = MinCostFlow(N*2+2)
s = N * 2
t = N * 2 + 1
for i, a in enumerate(A):
    mcf.add_edge(s, i, 1, 0)
    mcf.add_edge(N+i, t, 1, 0)
    for j in range(N):
        # 完全順列なので、A[i] == i のペアには辺を張れない
        if a == j+1:
            continue
        cost = a * abs(i - j)
        mcf.add_edge(i, N+j, 1, cost)

res = mcf.flow(s, t, N)
print(res)


```