Package for this problem was not updated by the problem writer or Codeforces administration after we’ve upgraded the judging servers. To adjust the time limit constraint, solution execution time will be multiplied by 2. For example, if your solution works for 400 ms on judging servers, then value 800 ms will be displayed and used to determine the verdict.

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

D. Not Quick Transformation

time limit per test

6 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLet *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* (*even*_{i} = *a*_{2i}, 1 ≤ 2*i* ≤ *n*), *odd* is an array consisting of the numberals whose numbers are odd in *а* (*odd*_{i} = *a*_{2i - 1}, 1 ≤ 2*i* - 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 *b*_{i}, such that *l* ≤ *i* ≤ *r* and *u* ≤ *b*_{i} ≤ *v*. You should print the query results modulo *mod*.

Input

The first line contains three integers *n*, *m*, *mod* (1 ≤ *n* ≤ 10^{18}, 1 ≤ *m* ≤ 10^{5}, 1 ≤ *mod* ≤ 10^{9}). Next *m* lines describe the queries. Each query is defined by four integers *l*, *r*, *u*, *v* (1 ≤ *l* ≤ *r* ≤ *n*, 1 ≤ *u* ≤ *v* ≤ 10^{18}).

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]

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Jan/20/2018 16:04:10 (d3).

Desktop version, switch to mobile version.

User lists

Name |
---|