Description

Edward gets a sequence of integers of length N, and he wants to find out how many of these "lucky"

Continuous sub-sequences. A continuous sub-sequence is called "lucky" when and only if the integer within that subsequence

And exactly the integer multiples of K. He asks you to write a program that calculates the number of successive subsequence sequences he likes.

Input

The input first line is an integer t, which indicates that there is a T group of data.

The first row of each group of data is two integers n (1 <= n <= 10^6), K (1 <= k <= 10^9).

The next line contains N integers Ai (| ai| <= 10^9).

Output

For each set of test data, the output line contains only one integer representing the successive sub-order Edward likes

The number of columns.

Sample Input

25 31 2 3 4 16 21 2 1 2 1 2

Sample Output

49

HINT

At first I did not think of the same remainder theorem, according to the Nanako of the great God, can be almost just card time and memory AC, but 鶸 does not.

Remember that after the simulation game, Nanako to the problem of the brain hole is almost completely close to the positive solution.

**Analysis:** According to the congruence theorem:(A-B)%c=a%c-b%c, you can think of using the prefix and, and record each prefix and the value after the modulo K,O (n^2) solution is a two-layer cyclic prefix array modulo k sets of formula, obviously timed out (good stupid , And then find several numbers with the same remainder. Example two: The prefix and the array after the modulo K is 1 1 0 0 1 1, find the remainder of the same b[0],b[1],b[4],b[5],4, then take the first and the end, the combination of different positions have (3+1) * (3)/2=6, the same, 0 have 2, However, note that the remainder of 0 means that the prefix of the position and itself is a multiple of k, so the combination of 0 is (2+1) *2/2=3.

**Note the point:** The number given may be negative, so the remainder needs to be transformed.

**1#include <stdio.h>2#include <iostream>3#include <string.h>4#include <algorithm>5 #definell Long Long6 using namespacestd;7 8 inta[1000010], b[1000010];9 intXe (inta)Ten { One return(A +1) * (a)/2; A } - intMain () - { the intT, T; - intN, K; - ll x, ans; -scanf"%d", &T); + while(t--) - { +scanf"%d%d", &n, &k); A atans=0; -scanf"%d", &b[0]); -b[0]%=K; - if(b[0] <0)//Note Negative Transform -b[0]= (b[0]+K)%K; - in - for(intI=1; i<n; i++) to { +scanf"%d",&a[i]); -b[i]=b[i-1]+A[i]; theb[i]%=K; * if(b[i]<0)//Note Negative Transform $b[i]= (b[i]+k)%K;Panax Notoginseng } -Sort (b, b+n);//easy handling after sorting theb[n]=-1; + /*for (int i=1; i<=n; i++) A { the cout<<b[i]< ""; + } - $ for (int i = 1; I <= n; i++)//o (n^2) obviously t Yes Brother $ { - For (int j = 0; J < i; j + +) - { the if (! ( (k+b[i]%k-b[j]%k)%k)) - {Wuyi ans++; the //printf ("%d~%d\n", i,j); - } Wu } - }*/ Aboutx=1; $ for(inti =1; I <= N; i++) - { - if(b[i]==b[i-1]) -X + +; A Else + { the if(b[i-1]!=0)//not 0 of the situation -x--; $Ans+=xe (x); thex=1; the } the } theprintf"%lld\n", ans); - } in}**

ACM Elective HUST1058 (city title) Lucky Sequence congruence theorem