#include<bits/stdc++.h>

using namespace std;

int main(void) {
    long n, d, a;
    cin >> n >> d >> a;
    
    vector<long> imos(2e5 + 1);
    vector<pair<long,long>> xh(n);
    for(auto &e: xh) cin >> e.first >> e.second;

    sort(xh.begin(), xh.end());

    auto is = xh.begin(), it = xh.begin();

    long r = 0;
    long s = 0;
    for(int i=0; i<n; i++) {
        s += imos[i];
        xh[i].second -= s;

        long t = max(0l, (xh[i].second + a - 1) / a * a);
        s += t;

        imos[upper_bound(xh.begin(), xh.end(), make_pair(xh[i].first + 2*d, (long)1e9)) - xh.begin()] += -t;

        r += t / a;
    }
    cout << r << endl;
}

### Prompt

Generate 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<bits/stdc++.h>

using namespace std;

int main(void) {
    long n, d, a;
    cin >> n >> d >> a;
    
    vector<long> imos(2e5 + 1);
    vector<pair<long,long>> xh(n);
    for(auto &e: xh) cin >> e.first >> e.second;

    sort(xh.begin(), xh.end());

    auto is = xh.begin(), it = xh.begin();

    long r = 0;
    long s = 0;
    for(int i=0; i<n; i++) {
        s += imos[i];
        xh[i].second -= s;

        long t = max(0l, (xh[i].second + a - 1) / a * a);
        s += t;

        imos[upper_bound(xh.begin(), xh.end(), make_pair(xh[i].first + 2*d, (long)1e9)) - xh.begin()] += -t;

        r += t / a;
    }
    cout << r << endl;
}

```