Kefa and Company, Binary Search

/**
 *    author:  compounding
 *    created: 2024-12-13 01:45:19
 **/
#include <bits/stdc++.h>
using namespace std;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
#define NeedForSpeed                  \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
#define f first
#define s second
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define endl "\n"
const int mod = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void solve()
{
    int n, d;
    cin >> n >> d;
    vpi a(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].f >> a[i].s;
    }
    sort(all(a));
    vi p(n + 1, 0);
    for (int i = 1; i <= n; i++)
    {
        p[i] = p[i - 1] + a[i - 1].s;
    }
    int mx = 0;
    for (int i = 0; i < n; i++)
    {
        // we check for each index, the maximum segment we can take such that the difference is less than d
        int l = i, r = n - 1, res = i;
        while (l <= r)
        {
            int mid = (l + r) / 2;
            if (a[mid].f - a[i].f < d)
            {
                res = mid;
                l = mid + 1;
            }
            else
            {
                r = mid - 1;
            }
        }
        mx = max(mx, p[res + 1] - p[i]);
    }
    cout << mx << endl;
    return;
}

signed main()
{
    NeedForSpeed;
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}