Not Quick Transformation
题目描述
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 α (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.
4 5 10000
2 3 4 5
2 4 1 3
1 2 2 4
2 3 3 5
1 3 3 4
0
5
3
3
3
2 5 10000
1 2 2 2
1 1 4 5
1 1 2 5
1 1 1 3
1 2 5 5
2
0
0
1
0
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]