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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 2010-2018 Mike Mirzayanov

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