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#5793

Cowmistry 1초 512MB

문제

Bessie has been procrastinating on her cow-mistry homework and now needs your help! She needs to create a mixture of three different cow-michals. As all good cows know though, some cow-michals cannot be mixed with each other or else they will cause an explosion. In particular, two cow-michals with labels a and b can only be present in the same mixture if a⊕b≤K(1≤K≤10^9).

NOTE: Here, a⊕b denotes the "bitwise exclusive or'' of non-negative integers a and b. This operation is equivalent to adding each corresponding pair of bits in base 2 and discarding the carry. For example,

0⊕0=1⊕1=0,

1⊕0=0⊕1=1,

5⊕7=101_2⊕111_2=010_2=2.

Bessie has N(1≤N≤2⋅10^4) boxes of cow-michals and the i-th box contains cow-michals labeled l_i through r_i inclusive (0≤l_i≤r_i≤10^9). No two boxes have any cow-michals in common. She wants to know how many unique mixtures of three different cow-michals she can create. Two mixtures are considered different if there is at least one cow-michal present in one but not the other. Since the answer may be very large, report it modulo 10^9+7.


입력

The first line contains two integers N and K.

Each of the next N lines contains two space-separated integers l_iand r_i. It is guaranteed that the boxes of cow-michals are provided in increasing order of their contents; namely, r_i<l_(i+1) for each 1≤i<N.


출력

The number of mixtures of three different cow-michals Bessie can create, modulo 10^9+7.


예제1

입력
1 13
0 199
출력
4280

We can split the chemicals into 13 groups that cannot cross-mix: (0…15), (16…31), …

(192…199). Each of the first twelve groups contributes 352 unique mixtures and the last contributes 56(since all (8 3) combinations of three different cow-michals from (192…199) are okay), for a total of 352⋅12+56=4280.


예제2

입력
6 147
1 35
48 103
125 127
154 190
195 235
240 250
출력
267188

출처

USACO 2020 December Platinum

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