codeforces 258D Little Elephant and Broken Sorting

D. Little Elephant and Broken Sorting

time limit per test

memory limit per test

input

output

The Little Elephant loves permutations of integers from 1 to n very much. But most of all he loves sorting them. To sort a permutation, the Little Elephant repeatedly swaps some elements. As a result, he must receive a permutation 1, 2, 3, ..., n.

This time the Little Elephant has permutation p₁, p₂, ..., p_n. Its sorting program needs to make exactly m moves, during the i-th move it swaps elements that are at that moment located at the a_i-th and the b_i-th positions. But the Little Elephant's sorting program happened to break down and now on every step it can equiprobably either do nothing or swap the required elements.

Now the Little Elephant doesn't even hope that the program will sort the permutation, but he still wonders: if he runs the program and gets some permutation, how much will the result of sorting resemble the sorted one? For that help the Little Elephant find the mathematical expectation of the number of permutation inversions after all moves of the program are completed.

We'll call a pair of integers i, j (1 ≤ i < j ≤ n) an inversion in permutatuon p₁, p₂, ..., p_n, if the following inequality holds: p_i > p_j.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1000, n > 1) — the permutation size and the number of moves. The second line contains n distinct integers, not exceeding n — the initial permutation. Next m lines each contain two integers: the i-th line contains integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the positions of elements that were changed during the i-th move.

Output

In the only line print a single real number — the answer to the problem. The answer will be considered correct if its relative or absolute error does not exceed 10^- 6.

Examples




    Input



2 1
1 2
1 2



    Output



0.500000000



    Input



4 3
1 3 2 4
1 2
2 3
1 4



    Output



3.000000000

【分析】

根本想不出来，但代码量巨短系列

¡ 一个 N 的排列， M 次操作，每次交换位置 i,j 的数，但是有一半概率不操作，问最终逆序对的期望个数。

¡ 用 P[i][j] 记录位置 i 比位置 j 大的概率

¡ 当交换 u,v 之后

¡
  P[u][v]=P[v][u]=0.5


¡
  p[u][k]=p[v][k]=0.5*(p[u][k]+p[v][k[)


¡
  p[k][u]=p[k][v]=0.5*(p[k][u]+p[k][v])

¡ 最后根据期望的定义暴力计算即可

【代码】

//codeforces 258D Little Elephant and Broken Sorting
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=1005;
int n,m;
int a[mxn];
double dp[mxn][mxn];
int main()
{
  double ans=0;
  int i,j,u,v;
  scanf("%d%d",&n,&m);
  fo(i,1,n) scanf("%d",&a[i]);
  fo(i,1,n)
    fo(j,i+1,n)
      if(a[i]>a[j]) dp[i][j]=1,dp[j][i]=0;
      else dp[i][j]=0,dp[j][i]=1;
  while(m--)
  {
    scanf("%d%d",&u,&v);
    dp[u][v]=dp[v][u]=0.5;
    fo(i,1,n) if(i!=u && i!=v)
      dp[u][i]=dp[v][i]=0.5*(dp[u][i]+dp[v][i]),
      dp[i][u]=dp[i][v]=0.5*(dp[i][u]+dp[i][v]);
  }
  fo(i,1,n) fo(j,i+1,n) ans+=dp[i][j];
  printf("%.9lf\n",ans);
  return 0;
}