NanoApe Loves Sequence

Accepts: 531

Submissions: 2481

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 262144/131072 K (Java/Others)

Problem Description

NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination!

In math class, NanoApe picked up sequences once again. He wrote down a sequence with nnn numbers on the paper and then randomly deleted a number in the sequence. After that, he calculated the maximum absolute value of the difference of each two adjacent remained numbers, denoted as FFF.

Now he wants to know the expected value of FFF, if he deleted each number with equal probability.

Input

The first line of the input contains an integer TTT, denoting the number of test cases.

In each test case, the first line of the input contains an integer nnn, denoting the length of the original sequence.

The second line of the input contains nnn integers A1,A2,...,AnA_1, A_2, ..., A_nA​1​​,A​2​​,...,A​n​​, denoting the elements of the sequence.

1≤T≤10, 3≤n≤100000, 1≤Ai≤1091 \le T \le 10,~3 \le n \le 100000,~1 \le A_i \le 10^91≤T≤10, 3≤n≤100000, 1≤A​i​​≤10​9​​

Output

For each test case, print a line with one integer, denoting the answer.

In order to prevent using float number, you should print the answer multiplied by nnn.

Sample Input

Copy

141 2 3 4

Sample Output

Copy

6

/*第二次来水BC，暑假最后一场了，这次收获不错，水出两题，虽然第二题不知道什么是线段树，但是还是按照自己想法搞出来了，加油*/ #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #define N 100010#define M 100000using namespace std;long long n, dp[N],pd[N],a[N], tep;long long Max(long long a,long long b,long long c){ long long d=max(a,b); long long e=max(b,c); return max(d,e);}int main(){ //freopen("in.txt","r",stdin); int t; scanf("%d",&t); while(t--) { memset(dp,0,sizeof(dp)); memset(pd,0,sizeof(pd)); vector 
              
                v; v.clear(); v.push_back(0); scanf("%lld",&n); for(int i = 1; i <= n; i++) { scanf("%lld",&tep); v.push_back(tep); } v.push_back(0); dp[1]=0; for(int i = 2; i <= n;i++) if(abs(v[i] - v[i - 1])>dp[i-1]) { dp[i]=abs(v[i] - v[i-1]); } else dp[i] = dp[i - 1]; reverse(v.begin(), v.end()); pd[1]=0; for(int i = 2; i <= n; i++) if(abs(v[i] - v[i-1]) > pd[i-1]) pd[i] = abs(v[i] - v[i-1]); else pd[i] = pd[i - 1]; reverse(v.begin(), v.end()); long long s = 0; for(int i=1;i<=n;i++) { if(i==1) s+=pd[n-1]; else if(i==n) s+=dp[n-1]; else s+=Max(dp[i-1],abs(v[i+1] - v[i-1]),pd[n-i]); } printf("%lld\n",s); } return 0;}