#include <cstdio>
#include <algorithm>
#include <utility>
#include <queue>

using namespace std;
typedef long long ll;

ll N, D, A, X, H, a, b, d, ans;
priority_queue<pair<ll, ll>, vector<pair<ll, ll> >, greater<pair<ll, ll> > > pq;

int main() {
  scanf("%lld %lld %lld", &N, &D, &A);
  for (ll i = 0; i < N; i++) {
    scanf("%lld %lld", &X, &H);
    pq.push(make_pair(X+D+1, (H+A-1)/A));
  }
  while (pq.size()) {
    a = pq.top().first;
    b = pq.top().second;
    pq.pop();
    if (b > d) {
      ans += b - d;
      pq.push(make_pair(a+2*D+1, d - b));
      d = b;
    }
    if (b < 0) d += b;
  }

  printf("%lld\n", ans);
}

### Prompt

Please formulate a CPP solution to the following problem:

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.

Examples

Input

3 3 2
1 2
5 4
9 2


Output

2


Input

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


Output

5


Input

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000


Output

3000000000

### Response

```cpp
#include <cstdio>
#include <algorithm>
#include <utility>
#include <queue>

using namespace std;
typedef long long ll;

ll N, D, A, X, H, a, b, d, ans;
priority_queue<pair<ll, ll>, vector<pair<ll, ll> >, greater<pair<ll, ll> > > pq;

int main() {
  scanf("%lld %lld %lld", &N, &D, &A);
  for (ll i = 0; i < N; i++) {
    scanf("%lld %lld", &X, &H);
    pq.push(make_pair(X+D+1, (H+A-1)/A));
  }
  while (pq.size()) {
    a = pq.top().first;
    b = pq.top().second;
    pq.pop();
    if (b > d) {
      ans += b - d;
      pq.push(make_pair(a+2*D+1, d - b));
      d = b;
    }
    if (b < 0) d += b;
  }

  printf("%lld\n", ans);
}

```