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 FOR(i,r,n) for(ll i=(ll)(r);i<(ll)(n);++i)
#define RFOR(i,r,n) for(ll i=(ll)(n-1);i>=r;--i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define RALL(x) x.rbegin(),x.rend()
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define pb(n) push_back(n)
#define endl '\n'
#define debug(x) cout << #x << " = " << (x) << endl
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c
#define VIN(V) for(int ii = 0; ii < (V).size(); ii++) {cin >> (V).at(ii);}
#define OUT(n) cout << n << endl
#define OUT2(n1,n2) cout << n1 << " " << n2 << endl
#define OUT3(n1,n2,n3) cout << n1 << " " << n2 << " " << n3 << endl
#define VOUT(V) REP(ii,(V).size()){cout << (V)[ii] << " ";} cout<<endl;
#define VOUT2(V) REP(jj,(V).size()){VOUT(V[jj])}
typedef long long int ll;
typedef vector<ll> vi;
typedef vector<vi > vvi;
typedef vector<pair<ll, ll> > vp;
typedef vector<string> vs;
typedef vector<char> vc;
typedef list<ll> lst;
typedef pair<ll, ll> P;
const long long INF = numeric_limits<ll>::max();
const long long MOD = 1000000007;
const long double PI = 3.1415926;
template <class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <class T> T &chmax(T &a, const T &b) { return a = max(a, b); }
ll n, m, k, ans, sum, cnt = 0; string s; char c; bool flg;

int main(){
  cin.tie(0);// ios::sync_with_stdio(false);
  ll d,a;
  IN3(n,d,a);
  vp xh;
  ll x,h;
  FOR(i,0,n) {
    scanf("%d %d", &x, &h);
    xh.push_back(P(x, h));
  }
  sort(xh.begin(), xh.end());
  FOR(i,0,n) {
    ll x = xh[i].first;
    ll h = xh[i].second;
    if (h<0) continue;

    cnt = h/a;
    if (h%a != 0) cnt++;
    ans += cnt;

    ll haba = x + d * 2;
    ll damage = cnt * a;
    FOR(j,i+1,n) {
      if (xh[j].first > haba) break;
      xh[j].second -= damage;
      if (xh[j].second < 0) xh[j].first = -1;
    }
  }
  OUT(ans);
}