// find the number of subarray whoose sum <=k and nums[i]>=0
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
 
int main() {
    ll n;
    cin>>n;
    ll k;cin>>k;
    ll b[n];
    for(ll i=0;i<n;i++){
        cin>>b[i];
    }ll count = 0 ;
    int sum = 0;
 
    // sort(b,b+n);
    for (int i = 0, j = 0; j < n; j++) {
        sum += b[j];
        while (sum > k ) {
            sum -= b[i++];
        }
        count += j-i+1;
    }
    cout<<count;
    return 0;
}//RRRRR

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5 
4
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