Problem:
Score : 600 points

Problem Statement

Silver Fox is fighting with N monsters.

The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.

Silver Fox can use bombs to attack the monsters.
Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A.
There is no way other than bombs to decrease the monster's health.

Silver Fox wins when all the monsters' healths become 0 or below.

Find the minimum number of bombs needed to win.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq D \leq 10^9

1 \leq A \leq 10^9

0 \leq X_i \leq 10^9

1 \leq H_i \leq 10^9

X_i are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N D A
X_1 H_1
:
X_N H_N

Output

Print the minimum number of bombs needed to win.

Sample Input 1

3 3 2
1 2
5 4
9 2

Sample Output 1

2

First, let us use a bomb at the coordinate 4 to decrease the first and second monsters' health by 2.

Then, use a bomb at the coordinate 6 to decrease the second and third monsters' health by 2.

Now, all the monsters' healths are 0.
We cannot make all the monsters' health drop to 0 or below with just one bomb.

Sample Input 2

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5

Sample Output 2

5

We should use five bombs at the coordinate 5.

Sample Input 3

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/* BISMILLAHIR RAHMANIR RAHIM */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main() {
    ios_base :: sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    ll n, d, A;
    cin >> n >> d >> A;
    vector< pair<ll, ll> > a(n);
    for(auto& i : a) {
        cin >> i.first >> i.second;
    } 
    sort(a.begin(), a.end());
    vector<ll> sg(n * 4, 0);
    function<void(int, int, int)> bld = [&](int nd, int l, int r) {
        if( l == r ) {
            sg[nd] = a[l].second;
            return;
        }
        int mid = l + (r - l) / 2, lft = nd << 1, rgt = lft | 1;
        bld( lft, l, mid );
        bld( rgt, mid + 1, r );
        sg[nd] = max(sg[lft], sg[rgt]);
    };
    function<ll(int, int, int, int, int)> qry = [&](int L, int R, int nd, int l, int r) {
        if( l > R or r < L or L > R ) return 0LL;
        if( l >= L and r <= R ) return sg[nd];
        int mid = l + (r - l) / 2, lft = nd << 1, rgt = lft | 1;
        return max(
            qry( L, R, lft, l, mid ),
            qry( L, R, rgt, mid + 1, r )
        );
    };
    bld( 1, 0, n - 1 );
    ll res = 0;
    for(int i = 0; i < n; i++) {
        int l = i, r = n - 1, id = l;
        while( l <= r ) {
            int mid = l + (r - l) / 2;
            if( a[i].first + d + d < a[mid].first ) r = mid - 1;
            else l = mid + 1, id = max(id, mid);
        }
        ll mx = qry( i, id, 1, 0, n - 1 );
        res += (mx / A + (mx % A > 0));
        i = id;
    }
    ll res1 = 0;
    for(int i = n - 1; i >= 0; i--) {
        int l = 0, r = i, id = r;
        while( l <= r ) {
            int mid = l + (r - l) / 2;
            if( a[i].first - d - d > a[mid].first ) l = mid + 1;
            else r = mid - 1, id = min(id, mid);
        }
        ll mx = qry( id, i, 1, 0, n - 1 );
        //cout << id << " " << i << " " << mx << "\n";
        res1 += (mx / A + (mx % A > 0));
        i = id;
    }
    cout << min(res, res1) << "\n";
}