D. Not Quick Transformation
divide and conquer; math
Let a be an array consisting of n numbers. The array's elements are numbered from 1 to n, even is an array consisting of the numerals whose numbers are even in a (eveni = a2i, 1 ≤ 2i ≤ n), odd is an array consisting of the numberals whose numbers are odd in а (oddi = a2i - 1, 1 ≤ 2i - 1 ≤ n). Then let's define the transformation of array F(a) in the following manner: if n > 1, F(a) = F(odd) + F(even), where operation "" + "" stands for the arrays' concatenation (joining together) if n = 1, F(a) = a Let a be an array consisting of n numbers 1, 2, 3, ..., n. Then b is the result of applying the transformation to the array a (so b = F(a)). You are given m queries (l, r, u, v). Your task is to find for each query the sum of numbers bi, such that l ≤ i ≤ r and u ≤ bi ≤ v. You should print the query results modulo mod.
The first line contains three integers n, m, mod (1 ≤ n ≤ 1018, 1 ≤ m ≤ 105, 1 ≤ mod ≤ 109). Next m lines describe the queries. Each query is defined by four integers l, r, u, v (1 ≤ l ≤ r ≤ n, 1 ≤ u ≤ v ≤ 1018).Please do not use the %lld specificator to read or write 64-bit integers in C++. Use %I64d specificator.
Print m lines each containing an integer — remainder modulo mod of the query result.
Let's consider the first example. First let's construct an array b = F(a) = F([1, 2, 3, 4]). Step 1. F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4]) Step 2. F([1, 3]) = F([1]) + F([3]) = [1] + [3] = [1, 3] Step 3. F([2, 4]) = F([2]) + F([4]) = [2] + [4] = [2, 4] Step 4. b = F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4]) = [1, 3] + [2, 4] = [1, 3, 2, 4] Thus b = [1, 3, 2, 4]. Let's consider the first query l = 2, r = 3, u = 4, v = 5. The second and third positions in the array b do not have numbers in the range [4, 5], so the sum obviously equals zero. Let's consider the second query l = 2, r = 4, u = 1, v = 3. The second and third positions have two numbers that belong to the range [1, 3], their sum equals 5.