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 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

66 4 2 7 2 732 3 61 3 41 1 6

output

24928

input

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

10155155521235

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;}