Cowmistry
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.
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 boxes of cow-michals and the i-th box contains cow-michals labeled l[i] through r[i] inclusive. 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.
Input
The first line contains two integers n (1 ≤ n ≤ 2 * 10^4) and k (1 ≤ k ≤ 10^9). Each of the next n lines contains two space-separated integers l[i] and r[i] (0 ≤ l[i] ≤ r[i] ≤ 10^9). 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.
Output
The number of mixtures of three different cow-michals Bessie can create, modulo 10^9 + 7.
Example
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 C3_8 combinations of three different cow-michals from (192...199) are okay), for a total of 352 * 12 + 56 = 4280.