codeforces689E: Mike and Geometry Problem

Description

Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's definef([l, r]) = r - l + 1 to be the number of integer points in the segment [l, r] with l ≤ r (say that ). You are given two integersn and k and n closed intervals [l_i, r_i] on OX axis and you have to find:

In other words, you should find the sum of the number of integer points in the intersection of any k of the segments.

As the answer may be very large, output it modulo 1000000007 (10⁹ + 7).

Mike can't solve this problem so he needs your help. You will help him, won't you?

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 200 000) — the number of segments and the number of segments in intersection groups respectively.

Then n lines follow, the i-th line contains two integers l_i, r_i ( - 10⁹ ≤ l_i ≤ r_i ≤ 10⁹), describing i-th segment bounds.

Output

Print one integer number — the answer to Mike's problem modulo 1000000007 (10⁹ + 7) in the only line.

Sample Input

Input

3 2
1 2
1 3
2 3

Output

5

Input

3 3
1 3
1 3
1 3

Output

3

Input

3 1
1 2
2 3
3 4

Output

6

Hint

In the first example:

;

;

.

So the answer is 2 + 1 + 2 = 5.

我们考虑每一个点的贡献

若一个点被i段区间覆盖

那么贡献就是C(k,i)

我们记录i=k到n的数量直接计算即可

对于线段统计我们可以打标记然后直接扫过去

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long mod=1000000007;
struct line
{
	int l;
	int x;
	bool operator< (line y) const
	{
		return l<y.l;
	}
}a[500001];
int ss[200001];
long long pow1[200001],pow2[200001];
inline long long power(long long x,int y)
{
	long long xt=1;
	while(y!=0)
	{
		if(y%2==1)
			xt=xt*x%mod;
		x=x*x%mod;
		y=y/2;
	}
	return xt;
}
inline long long prepare()
{
	long long i;
	pow1[0]=1;
	pow2[0]=1;
	for(i=1;i<=200000;i++)
	{
		pow1[i]=pow1[i-1]*i%mod;
		pow2[i]=power(pow1[i],mod-2);
	}
}
inline long long cale(long long k,long long n)
{
	return (pow1[n]*pow2[n-k]%mod)*pow2[k]%mod;
}
int main()
{
	prepare();
	int n,k;
	scanf("%d%d",&n,&k);
	int i;
	int p=0;
	for(i=1;i<=n;i++)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		p++;
		a[p].l=l;
		a[p].x=1;
		p++;
		a[p].l=r+1;
		a[p].x=-1;
	}
	sort(a+1,a+1+p);
	int d=1,sx=0,la=a[1].l;
	while(d<=p)
	{
		int sum=0;
		while(a[d].l==la)
		{
			sx+=a[d].x;
			d++;
		}
		ss[sx]+=a[d].l-la;
		la=a[d].l;
	}
	long long ans=0;
	for(i=k;i<=n;i++)
		ans=(ans+cale(k,i)*(long long)ss[i]%mod)%mod;
	printf("%I64d\n",ans);
	return 0;
}