CF689E：Mike and Geometry Problem（组合数）

E. Mike and Geometry Problem

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 

).
You are given two integers nand k and n closed
intervals [li, ri] 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 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

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 33 4

output

6

Note

In the first example:


;


;


.

So the answer is 2 + 1 + 2 = 5.

题意：给N个区间，问所有这N个区间组成的K元组区间的公共部分长度之和。

思路：每个点分开考虑，对于某个点被覆盖的次数>=k就贡献C(x, k)， 数的范围太大用map。

# include <bits/stdc++.h>
# define A first
# define B second
using namespace std;
typedef long long LL;
const int maxn = 2e5;
const LL mod = 1e9+7;
LL inv[maxn+30]={1,1}, fac[maxn+30]={1,1}, fi[maxn+30]= {1,1};
void init()
{
for(int i=2; i<=maxn; ++i)
{
fac[i] = i*fac[i-1]%mod;
inv[i] = (mod-mod/i)*inv[mod%i]%mod;
fi[i] = fi[i-1]*inv[i]%mod;
}
}
LL C(LL n, LL m)
{
return fac
*fi[m]%mod*fi[n-m]%mod;
}
map<int,int>m;
int main()
{
init();
int n, k, a, b;
scanf("%d%d",&n,&k);
for(int i=0; i<n; ++i)
{
scanf("%d%d",&a,&b);
++m[a];--m[b+1];
}
int l = m.begin()->A;
LL sum = 0, ans = 0;
for(auto it : m)
{
int dis = it.A - l;
if(sum >= k)
{
ans += C(sum, (LL)k)*dis;
ans %= mod;
}
sum += it.B;
l = it.A;
}
return 0*printf("%I64d\n",ans);;
}