Problem:
Score : 600 points

Problem Statement

We have a sequence of k numbers: d_0,d_1,...,d_{k - 1}.

Process the following q queries in order:

The i-th query contains three integers n_i, x_i, and m_i.
Let a_0,a_1,...,a_{n_i - 1} be the following sequence of n_i numbers: \begin{eqnarray} a_j = \begin{cases} x_i & ( j = 0 ) \\ a_{j - 1} + d_{(j - 1)~\textrm{mod}~k} & ( 0 < j \leq n_i - 1 ) \end{cases}\end{eqnarray}
Print the number of j~(0 \leq j < n_i - 1) such that (a_j~\textrm{mod}~m_i) < (a_{j + 1}~\textrm{mod}~m_i).

Here (y~\textrm{mod}~z) denotes the remainder of y divided by z, for two integers y and z~(z > 0).

Constraints

All values in input are integers.

1 \leq k, q \leq 5000

0 \leq d_i \leq 10^9

2 \leq n_i \leq 10^9

0 \leq x_i \leq 10^9

2 \leq m_i \leq 10^9

Input

Input is given from Standard Input in the following format:

k q
d_0 d_1 ... d_{k - 1}
n_1 x_1 m_1
n_2 x_2 m_2
:
n_q x_q m_q

Output

Print q lines.

The i-th line should contain the response to the i-th query.

Sample Input 1

3 1
3 1 4
5 3 2

Sample Output 1

1

For the first query, the sequence {a_j} will be 3,6,7,11,14.

(a_0~\textrm{mod}~2) > (a_1~\textrm{mod}~2)

(a_1~\textrm{mod}~2) < (a_2~\textrm{mod}~2)

(a_2~\textrm{mod}~2) = (a_3~\textrm{mod}~2)

(a_3~\textrm{mod}~2) > (a_4~\textrm{mod}~2)

Thus, the response to this query should be 1.

Sample Input 2

7 3
27 18 28 18 28 46 1000000000
1000000000 1 7
1000000000 2 10
1000000000 3 12

Sample Output 2

224489796
214285714
559523809

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#define FUP(i,x,y) for(int i=(x);i<=(y);i++)
#define FDW(i,x,y) for(int i=(x);i>=(y);i--)
#define MAXN 5010
#define INF 0x7fffffff
#define MOD 1000000007
#define ll long long
#define db double
using namespace std;
ll read(){
    ll w=0,flg=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')flg=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){w=w*10-'0'+ch,ch=getchar();}
    return w*flg;
}
ll k,q,n,x,m,D[MAXN],d[MAXN],cnt,sum;
void init()
{
	n=x=m=cnt=sum=0;
	memset(d,0,sizeof(d));
}
int main(){
	k=read(),q=read();
	FUP(i,1,k)
	{
		D[i]=read();
	}
	while(q--)
	{
		init();
		n=read(),x=read(),m=read();
		x%=m;
		FUP(i,1,k)
		{
			d[i]=D[i]%m;
			if(d[i]==0)
			{
				cnt++;
			}
			sum+=d[i];
		}
		cnt*=n/k,sum*=n/k;
		FUP(i,1,n%k-1)
		{
			if(d[i]==0)
			{
				cnt++;
			}
			sum+=d[i];
		}
		sum+=x;
		//printf("%lld,%lld,%lld\n",n,cnt,sum/m);
		printf("%lld\n",n-1-cnt-sum/m);
	}
    return 0;
}