## Algorithm

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:

1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her .
2. 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 typel 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
Copy
`66 4 2 7 2 732 3 61 3 41 1 6`
output
Copy
`24928`
input
Copy
`45 5 2 3101 2 42 1 41 1 12 1 42 1 21 1 11 3 31 1 31 4 41 2 2`
output
Copy
`10155155521235`
Note

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

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int const N = 1e5 + 1;
long long scum[N], uscum[N];
vector<int> arr;

int main() {
int n;
cin >> n;

arr.resize(n);
for(int i = 0; i < n; i++) cin >> arr[i];

uscum[0] = 0;
for(int i = 1; i <= n; i++)
uscum[i] = uscum[i-1] + arr[i-1];

sort(arr.begin(), arr.end());

scum[0] = 0;
for(int i = 1; i <= n; i++)
scum[i] = scum[i-1] + arr[i-1];

int q;
cin >> q;

int t, a, b;
while(q--) {
cin >> t >> a >> b;

if(t == 1)
cout << uscum[b] - uscum[a-1] << endl;
else
cout << scum[b] - scum[a-1] << endl;
}
}``````
Copy The Code &

Input

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

Output

cmd
24
9
28