Problem: K Average Ranges

Given the sequence a_1, a_2, .., a_N.

How many intervals in this sequence have an average value of K or more and a length of 1 or more?

input


N K
a_1 a_2 ... a_N


output

Output the answer.

Constraint

* 1 \ leq N \ leq 10 ^ 5
* 1 \ leq K \ leq 10 ^ 9
* 1 \ leq a_i \ leq 10 ^ 9



Input example


6 6
8 6 9 1 2 1


Output example


7






Example

Input

6 6
8 6 9 1 2 1


Output

7

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)a, i##_len=(b); i<i##_len; i++)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define int long long
using namespace std;
using P = pair<int, int>;
using ll = long long;
using prique = priority_queue<int>;
const ll INF = LLONG_MAX/3;

/*-----Library-----*/

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

struct UnionFind {
    vector<int> par; // 親ノード
    vector<int> rank; // ランク
    vector<int> sizeuf; //要素数
    UnionFind(int n = 1) {
        init(n);
    }

    void init(int n = 1) {
        par.resize(n); rank.resize(n); sizeuf.resize(n);
        for (int i = 0; i < n; ++i) {
            par[i] = i;
            rank[i] = 0;
            sizeuf[i]=1;
        }
    }

    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        sizeuf[x]+=sizeuf[y];
        return true;
    }

    int size(int x){
        return sizeuf[root(x)];
    }
};

long long gcd(long long a,long long b){
    if(b==0)return a;
    return gcd(b,a%b);
}

long long lcm(long long i,long long j){
    return i*(j/gcd(i,j));
}
//コンビネーション
/*
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 */
/*-----Library-----*/
const int MOD=1234567;
typedef long long int64;

template< class T >
struct BinaryIndexedTree
{
  vector< T > data;

  BinaryIndexedTree(int sz)
  {
    data.assign(++sz, 0);
  }

  T sum(int k)
  {
    T ret = 0;
    for(++k; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }

  void add(int k, T x)
  {
    for(++k; k < data.size(); k += k & -k) data[k] += x;
  }
};

signed main()
{
  int64 N, K, A[200000];

  cin >> N >> K;
  for(int i = 0; i < N; i++) cin >> A[i];

  int64 sum = 0;
  vector< int64 > odd;
  odd.push_back(0);
  for(int i = 0; i < N; i++) {
    sum += A[i];
    odd.push_back(sum - (i + 1) * K);
  }
  sort(begin(odd), end(odd));
  odd.erase(unique(begin(odd), end(odd)), end(odd));

  sum = 0;
  int64 ret = 0;
  BinaryIndexedTree< long long > bit(odd.size() + 2);
  bit.add(lower_bound(begin(odd), end(odd), 0) - begin(odd), 1);
  for(int i = 0; i < N; i++) {
    sum += A[i];
    ret += bit.sum(lower_bound(begin(odd), end(odd), sum - (i + 1) * K) - begin(odd));
    bit.add(lower_bound(begin(odd), end(odd), sum - (i + 1) * K) - begin(odd), 1);
  }
  cout << ret << endl;
}