codeforces#P689E. Mike and Geometry Problem

    ID: 28007 远端评测题 3000ms 256MiB 尝试: 0 已通过: 0 难度: 7 上传者: 标签>combinatoricsdata structuresdpgeometryimplementation*2000

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 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 n and 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?

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.

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

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.

Samples

3 2
1 2
1 3
2 3

5

3 3
1 3
1 3
1 3

3

3 1
1 2
2 3
3 4

6

Note

In the first example:

;

;

.

So the answer is 2 + 1 + 2 = 5.