For months Maxim has been coming to work on his favorite bicycle. And quite recently he decided that he is ready to take part in a cyclists’ competitions.

He knows that this year n competitions will take place. During the i-th competition the participant must as quickly as possible complete a ride along a straight line from point si to point fi (si < fi).

Measuring time is a complex process related to usage of a special sensor and a time counter. Think of the front wheel of a bicycle as a circle of radius r. Let’s neglect the thickness of a tire, the size of the sensor, and all physical effects. The sensor is placed on the rim of the wheel, that is, on some fixed point on a circle of radius r. After that the counter moves just like the chosen point of the circle, i.e. moves forward and rotates around the center of the circle.

At the beginning each participant can choose any point bi, such that his bike is fully behind the starting line, that is, bi < si - r. After that, he starts the movement, instantly accelerates to his maximum speed and at time tsi, when the coordinate of the sensor is equal to the coordinate of the start, the time counter starts. The cyclist makes a complete ride, moving with his maximum speed and at the moment the sensor’s coordinate is equal to the coordinate of the finish (moment of time tfi), the time counter deactivates and records the final time. Thus, the counter records that the participant made a complete ride in time tfi - tsi.

Maxim is good at math and he suspects that the total result doesn’t only depend on his maximum speed v, but also on his choice of the initial point bi. Now Maxim is asking you to calculate for each of n competitions the minimum possible time that can be measured by the time counter. The radius of the wheel of his bike is equal to r.

Input
The first line contains three integers n, r and v (1 ≤ n ≤ 100 000, 1 ≤ r, v ≤ 109) — the number of competitions, the radius of the front wheel of Max’s bike and his maximum speed, respectively.

Next n lines contain the descriptions of the contests. The i-th line contains two integers si and fi (1 ≤ si < fi ≤ 109) — the coordinate of the start and the coordinate of the finish on the i-th competition.

Output
Print n real numbers, the i-th number should be equal to the minimum possible time measured by the time counter. Your answer will be considered correct if its absolute or relative error will not exceed 10 - 6.

Namely: let’s assume that your answer equals a, and the answer of the jury is b. The checker program will consider your answer correct if .

Example
Input
2 1 2
1 10
5 9
Output
3.849644710502
1.106060157705

題意大概就是給出自行車比賽的計時方式，輪胎上的計時器與起點線位於豎直面上開始即時，與終點線位於同一豎直面上時停止計時，所以如圖所示。

開始計時的時候，s點到輪胎圓心的水平距離差和停止計時時的距離差就是不參與計時的路程，我們就是需要找到一種策略使得這段不參與計時的路程達到最長，易知（不會證）如圖上那種策略時最優的，也就是開始計時時和停止計時的時候計時器是對稱的，假設總路長為dist，我們取len=dist減去路程中完全圓周長的長度（dist/(周長)*周長），剩下的就是最後需要走的距離l和節省下來的兩段距離s。這裡剩下來的距離S等於開始時計時器距離圓心的水平距離*2,所以只要我們找到滿足l+s=len的最小的l就行了，這裡可知s=r*sin（l / r) ，所以l+s是單調的，我們只要用二分法就能快速找到最小的滿足條件的l了。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
int n, r, v,s,f;
double pai = acos(-1);
int main(){
    scanf("%d%d%d", &n, &r, &v);
    for (int i = 0; i < n; i++){
        scanf("%d%d", &s, &f);
        double len = f - s;
        double c = pai*r * 2;
        int k = len / c;
        double ans = k*c / v;
        len -= c*k;
        len /= 2;
        double lef = 0, rig = len;
        for (int i = 0; i < 50; i++){
            double mid = (lef + rig) / 2;
            if (mid + r*sin(mid / r) >= len)rig = mid;
            else lef = mid;
        }
        ans += 2 * lef / v;
        printf("%.15lf\n", ans);
    }
    return 0;
}