#P689E. Mike and Geometry Problem

    ID: 3580 Type: RemoteJudge 3000ms 256MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>combinatoricsdata structuresdpgeometryimplementation*2000

Mike and Geometry Problem

No submission language available for this 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.