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>
using namespace std;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000000000000
#define MOD 1000000007
#define all(v) v.begin(),v.end()
#define pb push_back
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int kaijo[2000010];
struct edge{int to,cost;};
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
void init_fact(){
    kaijo[0]=1;
    for(int i=1;i<=2000000;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=MOD;
    }
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%MOD;
    int memo=modpow(a,b/2);
    return memo*memo%MOD;
}
int comb(int a,int b){
    if(!kaijo[0])init_fact();
    return kaijo[a]*modpow(kaijo[a-b],MOD-2)%MOD*modpow(kaijo[b],MOD-2)%MOD;
}
int inv(int x){
    x=modpow(x,MOD-2);
    return x;
}
bool kosa(double ax,double ay,double bx,double by,double cx,double cy,double dx,double dy){
    double ta=(cx-dx)*(ay-cy)+(cy-dy)*(cx-ax);
    double tb=(cx-dx)*(by-cy)+(cy-dy)*(cx-bx);
    double tc=(ax-bx)*(cy-ay)+(ay-by)*(ax-cx);
    double td=(ax-bx)*(dy-ay)+(ay-by)*(ax-dx);
    return tc*td<0&&ta*tb<0;
}
double dist(double ax,double ay,double bx,double by){
    return sqrt((bx-ax)*(bx-ax)+(ay-by)*(ay-by));
}
int n,k,a[100010],b[100010],ans;
map<int,int>mp;
int bit[100010];//[1,N+1]
int sum(int i){
    int s=0;
    while(i>0){
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}
void add(int i,int x){
    while(i<=n){
        bit[i]+=x;
        i+=i&-i;
    }
}
signed main(){
    cin>>n>>k;
    rep(i,n)cin>>a[i];
    rep(j,n+1){
        b[j]=b[j-1]+a[j-1]-k;
    }
    rep(i,n+1)mp[b[i]]=1;
    int now=0;
    for(P p:mp){
        if(p.se){
            mp[p.fi]=now;
            now++;
        }
    }
    rep(i,n+1){
        b[i]=mp[b[i]];
    }
    n++;
    rep(i,n){
        ans+=sum(b[i]+1);
        add(b[i]+1,1);
    }
    cout<<ans<<endl;
    return 0;
}