As you know, the most intelligent beings on the Earth are, of course, cows. This conclusion was reached long ago by the Martian aliens, as well as a number of other intelligent civilizations from outer space.

Sometimes cows gather into cowavans. This seems to be seasonal. But at this time the cows become passive and react poorly to external stimuli. A cowavan is a perfect target for the Martian scientific saucer, it's time for large-scale abductions, or, as the Martians say, raids. Simply put, a cowavan is a set of cows in a row.

If we number all cows in the cowavan with positive integers from 1 to n, then we can formalize the popular model of abduction, known as the (a, b)-Cowavan Raid: first they steal a cow number a, then number a + b, then — number a + 2·b, and so on, until the number of an abducted cow exceeds n. During one raid the cows are not renumbered.

The aliens would be happy to place all the cows on board of their hospitable ship, but unfortunately, the amount of cargo space is very, very limited. The researchers, knowing the mass of each cow in the cowavan, made p scenarios of the (a, b)-raid. Now they want to identify the following thing for each scenario individually: what total mass of pure beef will get on board of the ship. All the scenarios are independent, in the process of performing the calculations the cows are not being stolen.

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 400001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
inline void P(int x)
{Num=0;if(!x){putchar('0');puts("");return;}while(x>0)CH[++Num]=x%10,x/=10;while(Num)putchar(CH[Num--]+48);puts("");
}
//**************************************************************************************

ll a[maxn],n,m;
int cx[maxn],cy[maxn];
ll ans[maxn];
ll dp[maxn];
int kiss=600;
vector<int> d[600];
int main()
{//test;n=read();for(int i=1;i<=n;i++)a[i]=read();int m=read();for(int i=1;i<=m;i++){cx[i]=read(),cy[i]=read();if(cy[i]<kiss)d[cy[i]].push_back(i);else{ll ans1=0;for(int j=cx[i];j<=n;j+=cy[i]){ans1+=a[j];}ans[i]=ans1;}}for(int i=1;i<kiss;i++){if(d[i].size()){for(int j=n;j;j--){if(j+i>n)dp[j]=a[j];elsedp[j]=dp[j+i]+a[j];}for(int j=0;j<d[i].size();j++)ans[d[i][j]]=dp[cx[d[i][j]]];}}for(int i=1;i<=m;i++)cout<<ans[i]<<endl;
}