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 \le 2i \le n), odd is an array consisting of the numberals whose numbers are odd in \alpha (oddi=a2i-1, 1 \le 2i-1 \le n). Then let's define the transformation of array F(a) in the following manner:
if n \gt 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 \le i \le r and u \le bi \le v. You should print the query results modulo mod.

The first line contains three integers n, m, mod (1 \le n \le 1018,1 \le m \le 105,1 \le mod \le 109). Next m lines describe the queries. Each query is defined by four integers l, r, u, v (1 \le l \le r \le n, 1 \le u \le v \le 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.