Problem:
Score : 600 points

Problem Statement

You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.

a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints

All input values are integers.

1 ≤ N ≤ 2 \times 10^5

1 ≤ K ≤ 10^9

1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:

N K
a_1
a_2
:
a_N

Output

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

Sample Input 1

3 6
7
5
7

Sample Output 1

5

All the non-empty contiguous subsequences of a are listed below:

{a_1} = {7}

{a_1, a_2} = {7, 5}

{a_1, a_2, a_3} = {7, 5, 7}

{a_2} = {5}

{a_2, a_3} = {5, 7}

{a_3} = {7}

Their means are 7, 6, 19/3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a_1} and {a_3} are indistinguishable by the values of their elements, but we count them individually.

Sample Input 2

1 2
1

Sample Output 2

0

Sample Input 3

7 26
10
20
30
40
30
20
10

Sample Output 3

13

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
int A[999999],W[999999];
long long M(int l,int r){
	if(l+1>=r)return 0;
	if(l+2==r){
		if(A[l]<=A[l+1])return 0;
		int z=A[l];A[l]=A[l+1];A[l+1]=z;
		return 1;
	}
	int m = (l+r)/2,i=l, j=m, k=l;
	long long cl=M(l, m),cr=M(m,r),c=0;
	while(i<m&&j<r){
		if(A[i]<=A[j])W[k++]=A[i++];
		else{
			W[k++]=A[j++];
			c+=m-i;
		}
	}
	while(i<m)W[k++]=A[i++];
	while(j<r)W[k++]=A[j++];
	for(i=l;i<r;i++)A[i]=W[i];
	return cl+cr+c;
}
int main(){
	int T,i,N;
	//PKU2299,aizu0167: バブルソートの交換回数
	//for(;scanf("%d",&N),N;printf("%lld\n",M(0,N)))for(i=0;i<N;i++)scanf("%d",A+i);
	//hackerrank~runningtime,aizuALDS1~5D
	//scanf("%d",&N);for(i=0;i<N;i++)scanf("%d",A+i);printf("%lld\n",M(0,N));
	//hackerrank~30-sorting
	//scanf("%d",&N);for(i=0;i<N;i++)scanf("%d",A+i);printf("Array is sorted in %lld swaps.\n",M(0,N)),printf("First Element: %d\nLast Element: %d\n",A[0],A[N-1]);
	//spojINVCNT,hackerrank~insertionsort
	//for(scanf("%d",&T);T--;printf("%lld\n",M(0,N)))for(scanf("%d",&N),i=0;i<N;++i)scanf("%d",A+i);
	//codeiq4322: 光線の交差数
	//for(N=0;~scanf("%d",A+N);N++);printf("%lld\n",M(0,N));
	//kcs1009K: Nパズルの可解性
	//for(i=0;~(N=getchar());)if('0'<N&&N<'9')A[i++]=N-'0';
	//puts(M(0,i)%2?"lmpossible":"Possible"); //"l"? lol!
	//atcoderutpc2014F: パリティ
	//for(scanf("%d",&T),getchar();T--;printf("%lld\n",M(0,N)%2))for(N=0;~(A[N]=getchar())&&A[N]!='\n';N++);
	//kcs1031A
	//scanf("%d%d",&N,&T);for(i=0;i<N;i++)scanf("%d",A+i);N=M(0,N)-T;printf("%d\n",N>0?N:0);
	//atcoderARC075E: 算術平均がT以上
	scanf("%d%d",&N,&T);for(i=1;i<=N;i++)scanf("%d",&A[i]);for(i=1;i<=N;i++)A[i]+=A[i-1]-T;printf("%lld\n",(long long)N*-~N/2-M(0,N+1));
	return 0;
}