Wet Shark and Flowers（思维）

There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i and i + 1 are neighbours for all i from 1 to n - 1. Sharks n and 1 are neighbours too.

Each shark will grow some number of flowers s_i. For i-th shark value s_i is random integer equiprobably chosen in range from l_i to r_i. Wet Shark has it's favourite prime number p, and he really likes it! If for any pair of neighbouring sharks iand j the product s_i·s_j is divisible by p, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.

At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.

Input

The first line of the input contains two space-separated integers n and p (3 ≤ n ≤ 100 000, 2 ≤ p ≤ 10⁹) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that p is prime.

The i-th of the following n lines contains information about i-th shark — two space-separated integers l_i and r_i(1 ≤ l_i ≤ r_i ≤ 10⁹), the range of flowers shark i can produce. Remember that s_i is chosen equiprobably among all integers from l_i to r_i, inclusive.

Output

Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10^- 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $\text{[math]}$ .

Sample test(s)

input

3 2 1 2 420 421 420420 420421

output

4500.0

input

3 5 1 4 2 3 11 14

output

0.0

Note

A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.

Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (s₀, s₁, s₂) each shark grows:

(1, 420, 420420): note that s₀·s₁ = 420, s₁·s₂ = 176576400, and s₂·s₀ = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.
(1, 420, 420421): now, the product s₂·s₀ is not divisible by 2. Therefore, sharks s₀ and s₂ will receive 1000 dollars, while shark s₁ will receive 2000. The total is 4000.
(1, 421, 420420): total is 4000
(1, 421, 420421): total is 0.
(2, 420, 420420): total is 6000.
(2, 420, 420421): total is 6000.
(2, 421, 420420): total is 6000.
(2, 421, 420421): total is 4000.

The expected value is $\text{[math]}$ .

In the second sample, no combination of quantities will garner the sharks any money.

题解：这个题竟然没做出来。。。。都怪自己想的太复杂了，哪有那么复杂，对于每两个相邻的数找到出现质数的倍数出现的概率，其中每个质数倍数出现的次数是(r[i]/q-(l[i]-1)/q)；以前都做过这类题。。。对于每两个概率乘以2000，加和就好了啊；；；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%I64d",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%I64d",x)
#define P_ printf(" ")
typedef __int64 LL;
const int MAXN=100010;
int l[MAXN],r[MAXN];
int main(){
	int n,q;
	while(~scanf("%d%d",&n,&q)){
		for(int i=0;i<n;i++){
			scanf("%d%d",&l[i],&r[i]);
		}
		l[n]=l[0];r[n]=r[0];
		double ans=0;
		for(int i=0;i<n;i++){
			int n1=(r[i]/q-(l[i]-1)/q),n2=(r[i+1]/q-(l[i+1]-1)/q);
			ans+=2000.0*(1.0-1.0*(r[i]-l[i]+1-n1)*(r[i+1]-l[i+1]+1-n2)/(r[i]-l[i]+1)/(r[i+1]-l[i+1]+1));
		}
		printf("%.10f
",ans);
	}
	return 0;
	
}