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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("-1"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
#define bn(x) ((1<<x)-1)
#define newline puts("")
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,const vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;

// Binary Indexed Tree
struct bit {
  typedef int TT;
  int n; vector<TT> d;
  bit() {}
  bit(int mx): n(mx), d(mx) {}
  void add(int i, TT x=1) {
    for (++i;i<n;i+=i&-i) d[i] += x;
  }
  TT sum(int i) {
    TT x = 0;
    for (++i;i;i-=i&-i) x += d[i];
    return x;
  }
};
//

// coordinate compression
struct X {
  typedef int T;
  vector<T> d;
  void add(T x) { d.pb(x);}
  void init() {
    sort(rng(d));
    d.erase(unique(rng(d)), d.end());
  }
  int size() const { return sz(d);}
  int operator[](int i) const { return d[i];}
  int operator()(T x) const { return upper_bound(rng(d),x)-d.begin()-1;}
};
//



int main() {
  int n,x;
  scanf("%d%d",&n,&x);
  vl a(n);
  cin>>a;
  rep(i,n) a[i] -= x;
  rep(i,n-1) a[i+1] += a[i];
  X xs;
  rep(i,n) xs.add(a[i]);
  xs.add(0);
  xs.init();
  bit d(sz(xs)+1);
  d.add(xs(0));
  ll ans = 0;
  rep(i,n) {
    ans += d.sum(xs(a[i]));
    d.add(xs(a[i]));
  }
  cout<<ans<<endl;
  return 0;
}