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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

from bisect import bisect_right
N, D, A = map(int, input().split())

INF = 2**31-1

LV = (N-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [0]*(2*N0)

def gindex(l, r):
    L = (l + N0) >> 1; R = (r + N0) >> 1
    lc = 0 if l & 1 else (L & -L).bit_length()
    rc = 0 if r & 1 else (R & -R).bit_length()
    for i in range(LV):
        if rc <= i:
            yield R
        if L < R and lc <= i:
            yield L
        L >>= 1; R >>= 1

# 遅延伝搬処理
def propagates(*ids):
    for i in reversed(ids):
        v = lazy[i-1]
        if not v:
            continue
        lazy[2*i-1] += v; lazy[2*i] += v
        data[2*i-1] += v; data[2*i] += v
        lazy[i-1] = 0

# 区間[l, r)にxを加算
def update(l, r, x):
    *ids, = gindex(l, r)
    propagates(*ids)

    L = N0 + l; R = N0 + r
    while L < R:
        if R & 1:
            R -= 1
            lazy[R-1] += x; data[R-1] += x
        if L & 1:
            lazy[L-1] += x; data[L-1] += x
            L += 1
        L >>= 1; R >>= 1
    for i in ids:
        data[i-1] = max(data[2*i-1], data[2*i])

# 区間[l, r)内の最小値を求める
def query(l, r):
    propagates(*gindex(l, r))
    L = N0 + l; R = N0 + r

    s = -INF
    while L < R:
        if R & 1:
            R -= 1
            s = max(s, data[R-1])
        if L & 1:
            s = max(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s


def solve():
    m, l = zip(*sorted(tuple(map(int, input().split())) for _ in range(N)))
    for i, h in enumerate(l):
        update(i, i+1, h)

    ans = 0
    for i in range(N):
        v = query(i, i+1)
        if v>0:
            c = (v+A-1)//A
            ans += c
            pos = bisect_right(m, m[i]+2*D)
            update(i, pos, -c*A)

    print(ans)


if __name__ == "__main__":
    solve()