Problem4075--Not Quick Transformation

4075: Not Quick Transformation

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Time Limit : 2.000 sec  Memory Limit : 256 MiB

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Description

time limit per test
6 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Input

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.

Output

Print m lines each containing an integer − remainder modulo mod of the query result.

Examples
Input
4 5 10000
2 3 4 5
2 4 1 3
1 2 2 4
2 3 3 5
1 3 3 4
Output
0
5
3
3
3
Input
2 5 10000
1 2 2 2
1 1 4 5
1 1 2 5
1 1 1 3
1 2 5 5
Output
2
0
0
1
0
Note

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.

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