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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define INF (1e9)
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
const long double PI = acos(-1.0L);
const long long MOD = 1000000007LL;
// const long long MOD = 998244353LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true;} return false; }
////////////////////////////////////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    ll N,D,A; cin >> N >> D >> A;
    vector<ll> X(N), H(N);
    rep(i,N) cin >> X[i] >> H[i];

        auto B = H;
        vector<ll> cnt(N+1);
        rep(i,N) {
            B[i] = (B[i]+A-1LL) / A;
        }
        ll k = 0;
        for (ll i = 0; i < N; i++) {
            k += max(0LL,B[i]-cnt[i]+cnt[max(0LL,i-2*D)]);
            cnt[i+1] = k;
        }

        cout << k << endl;
}