Hongjun's Intersection

There are $N$ one dimensional segments on a horizontal line. Call the $i$-th segment $[L_i, R_i]$ ($L_i \le R_i$). The length of a segment is $f([L_i, R_i]) = R_i - L_i + 1$, and the length of the empty set is $f(\varnothing) = 0$.

You are given a positive integer $k$ with $k \le N$. For every way of choosing $k$ distinct segments, add up the length of the intersection of the chosen segments:

$\sum_{1 \le i_1 < i_2 < \cdots < i_k \le N} f\bigl([L_{i_1}, R_{i_1}] \cap [L_{i_2}, R_{i_2}] \cap \cdots \cap [L_{i_k}, R_{i_k}]\bigr)$

Write a program that computes this sum.

The answer can be very large, so print it modulo $10^9 + 7$.

The first line contains $N$ and $k$. ($1 \le k \le N \le 200{,}000$)

Each of the next $N$ lines contains two integers $L_i$ and $R_i$ describing the $i$-th segment. ($-10^9 \le L_i \le R_i \le 10^9$)

Print on one line the sum of the intersection lengths over every way of choosing $k$ distinct segments, modulo $10^9 + 7$.

When the segments are $[1, 2]$, $[1, 3]$, $[2, 3]$ and $k = 2$, there are three ways to choose, and the intersection lengths are:

$f([1,2] \cap [1,3]) = f([1,2]) = 2$

$f([1,2] \cap [2,3]) = f([2,2]) = 1$

$f([1,3] \cap [2,3]) = f([2,3]) = 2$

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