B. Kuriyama Mirai's Stones

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n.
The cost of the i-th stone is vi.
Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:

She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n),
and you should tell her 

.

Let ui be
the cost of the i-th cheapest stone (the cost that will be on the i-th
place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n),
and you should tell her 

.

For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.

Input

The first line contains an integer n (1 ≤ n ≤ 105).
The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) —
costs of the stones.

The third line contains an integer m (1 ≤ m ≤ 105) —
the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2),
describing a question. If type equal to 1,
then you should output the answer for the first question, else you should output the answer for the second one.

Output

Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the
order of input.

Examples

input

6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6

output

24
9
28

input

4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2

output

10
15
5
15
5
5
2
12
3
5

Note

Please note that the answers to the questions may overflow 32-bit integer type.

解题说明：此题是一道模拟题，按照题目要求输出给定区间的数字之和，或者对数列进行排序，输出给定区间之和。

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include <map>
using namespace std;

int main()
{
int n,q,t,l,r,i;
long long int a[100010],b[100010],c[100010];
cin>>n;
for( i=0; i<n; i++)
{
cin>>a[i];
b[i+1] = a[i] + b[i];
}
sort(a,a+n);
for( i=0; i<n; i++)
{
c[i+1] = a[i] + c[i];
}
cin>>q;
while(q--)
{
cin>>t>>l>>r;
if(t==1)
{
cout<<b[r]-b[l-1]<<endl;
}
else
{
cout<<c[r]-c[l-1]<<endl;
}
}
return 0;
}