Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 【逆元求组合数 && 离散化】

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本文介绍了一道关于计算多个区间交集长度总和的问题，通过离散化和使用组合数学的方法，提供了一个高效的解决方案。

任意门：http://codeforces.com/contest/689/problem/E

Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define f([l, r]) = r - l + 1 to be the number of integer points in the segment [l, r] with l ≤ r (say that $\text{[math]}$ ). You are given two integers nand k and n closed intervals [li, ri] on OX axis and you have to find:

$\text{[math]}$

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 li, ri ( - 109 ≤ li ≤ ri ≤ 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.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

input

Copy

output

Copy

Note

In the first example:

$\text{[math]}$ ;

$\text{[math]}$ .

So the answer is 2 + 1 + 2 = 5.

大概题意：

有 N 个区间，从其中取 K 个区间。所以有 C（N， K）种组合，求每种组合区间交集长度的总和。

解题思路：

丢开区间的角度，从每个结点的角度来看，其实每个结点的贡献是 C（cnt， K） cnt 为该结点出现的次数，所以只要O（N）扫一遍统计每个结点的贡献就是答案。

思路清晰，但考虑到数据的规模，这里需要注意和需要用到两个技巧：

一是离散化，这里STL里的 vector 和 pair 结合用，结合区间加法的思想进行离散化。

二是求组合数时除数太大，考虑到精度问题需要用逆元来计算。

AC code：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 2e5+7;
 4 const int mod = 1e9+7;
 5 long long fac[maxn];
 6 
 7 long long qpow(long long a,long long b)  //快速幂
 8 {
 9     long long ans=1;a%=mod;
10     for(long long i=b;i;i>>=1,a=a*a%mod)
11         if(i&1)ans=ans*a%mod;
12     return ans;
13 }
14 
15 long long C(long long n,long long m)    //计算组合数
16 {
17     if(m>n||m<0)return 0;
18     long long s1=fac[n], s2=fac[n-m]*fac[m]%mod;   //除数太大，逆元处理
19     return s1*qpow(s2,mod-2)%mod;
20 }
21 int n,k;
22 int l[maxn],r[maxn];      //左端点， 右端点
23 int main()
24 {
25     fac[0]=1;
26     for(int i=1;i<maxn;i++)     //预处理全排列
27         fac[i]=fac[i-1]*i%mod;
28 
29     scanf("%d%d",&n,&k);
30     for(int i=1;i<=n;i++){
31         scanf("%d",&l[i]);
32         scanf("%d",&r[i]);
33     }
34     vector<pair<int,int> >op;
35     for(int i=1;i<=n;i++){                //离散化
36         op.push_back(make_pair(l[i]-1,1));  //区间加法标记
37         op.push_back(make_pair(r[i],-1));
38     }
39     sort(op.begin(),op.end());  //升序排序
40     long long ans = 0;          //初始化
41     int cnt=0;
42     int la=-2e9;
43     for(int i=0;i<op.size();i++){                //计算每点的贡献
44         ans=(ans+C(cnt,k)*(op[i].first-la))%mod;
45         la=op[i].first;
46         cnt+=op[i].second;                    //该点的前缀和就是该点的出现次数
47     }
48     cout<<ans<<endl;
49 }