Codeforces 989D A Shade of Moonlight 二分+數學

D. A Shade of Moonlight

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The sky can be seen as a one-dimensional axis. The moon is at the origin whose coordinate is 00.

There are nn clouds floating in the sky. Each cloud has the same length ll. The ii-th initially covers the range of (xi,xi+l)(xi,xi+l) (endpoints excluded). Initially, it moves at a velocity of vivi, which equals either 11 or −1−1.

Furthermore, no pair of clouds intersect initially, that is, for all 1≤i<j≤n1≤i<j≤n, |xi−xj|≥l|xi−xj|≥l.

With a wind velocity of ww, the velocity of the ii-th cloud becomes vi+wvi+w. That is, its coordinate increases by vi+wvi+w during each unit of time. Note that the wind can be strong and clouds can change their direction.

You are to help Mino count the number of pairs (i,j)(i,j) (i<ji<j), such that with a proper choice of wind velocity ww not exceeding wmaxwmax in absolute value (possibly negative and/or fractional), the ii-th and jj-th clouds both cover the moon at the same future moment. This wwdoesn't need to be the same across different pairs.

Input

The first line contains three space-separated integers nn, ll, and wmaxwmax (1≤n≤1051≤n≤105, 1≤l,wmax≤1081≤l,wmax≤108) — the number of clouds, the length of each cloud and the maximum wind speed, respectively.

The ii-th of the following nn lines contains two space-separated integers xixi and vivi (−108≤xi≤108−108≤xi≤108, vi∈{−1,1}vi∈{−1,1}) — the initial position and the velocity of the ii-th cloud, respectively.

The input guarantees that for all 1≤i<j≤n1≤i<j≤n, |xi−xj|≥l|xi−xj|≥l.

Output

Output one integer — the number of unordered pairs of clouds such that it's possible that clouds from each pair cover the moon at the same future moment with a proper choice of wind velocity ww.

Examples

input

Copy

5 1 2
-2 1
2 1
3 -1
5 -1
7 -1

output

Copy

4

input

Copy

4 10 1
-20 1
-10 -1
0 1
10 -1

output

Copy

1

Note

In the first example, the initial positions and velocities of clouds are illustrated below.

The pairs are:

• (1,3)(1,3), covering the moon at time 2.52.5 with w=−0.4w=−0.4;
• (1,4)(1,4), covering the moon at time 3.53.5 with w=−0.6w=−0.6;
• (1,5)(1,5), covering the moon at time 4.54.5 with w=−0.7w=−0.7;
• (2,5)(2,5), covering the moon at time 2.52.5 with w=−2w=−2.

Below is the positions of clouds at time 2.52.5 with w=−0.4w=−0.4. At this moment, the 11-st and 33-rd clouds both cover the moon.

In the second example, the only pair is (1,4)(1,4), covering the moon at time 1515 with w=0w=0.

Note that all the times and wind velocities given above are just examples among infinitely many choices.

思路上大體上就是二分了，顯然只有初始速度方向相反的雲才能相遇，所以猜測應該就是將方向不同的雲記錄在兩個數組裏，然後枚舉其中一個，二分另一個。然而應該用什麼指標來判斷雲是否能相遇呢？到這兒騷操作就來了，我們想象沒有風在吹，月亮自己在跑，如下圖。

橙色代表月亮的軌跡，藍色代表雲的軌跡。如果兩個藍條能在橙色部分相遇，那麼就說明有一對雲是符合要求的。根據w必定大於1，可知橙色線和橫軸夾角不得大於45°。那就保證藍條相交得到的菱形，上面的頂點在橙色區域裏，因此xj−xi+l2>|xj−xi+l2+xi|/w。

從第一個滿足這個條件的雲開始，往後的雲都滿足這個條件，我們把它加在結果裏。

代碼奇醜無比，不貼了。