Turtle vs
Description
Isaac begins his training. There are $n$ running tracks available, and the $i$-th track ($1 \le i \le n$) consists of $a_i$ equal-length sections.
Given an integer $u$ ($1 \le u \le 10^9$), finishing each section can increase Isaac's ability by a certain value, described as follows:
- Finishing the $1$-st section increases Isaac's performance by $u$.
- Finishing the $2$-nd section increases Isaac's performance by $u-1$.
- Finishing the $3$-rd section increases Isaac's performance by $u-2$.
- $\ldots$
- Finishing the $k$-th section ($k \ge 1$) increases Isaac's performance by $u+1-k$. (The value $u+1-k$ can be negative, which means finishing an extra section decreases Isaac's performance.)
You are also given an integer $l$. You must choose an integer $r$ such that $l \le r \le n$ and Isaac will finish each section of each track $l, l + 1, \dots, r$ (that is, a total of $\sum_{i=l}^r a_i = a_l + a_{l+1} + \ldots + a_r$ sections).
Answer the following question: what is the optimal $r$ you can choose that the increase in Isaac's performance is maximum possible?
If there are multiple $r$ that maximize the increase in Isaac's performance, output the smallest $r$.
To increase the difficulty, you need to answer the question for $q$ different values of $l$ and $u$.
The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The descriptions of the test cases follow.
The first line contains a single integer $n$ ($1 \le n \le 10^5$).
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^4$).
The third line contains a single integer $q$ ($1 \le q \le 10^5$).
The next $q$ lines each contain two integers $l$ and $u$ ($1 \le l \le n, 1 \le u \le 10^9$) — the descriptions to each query.
The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. The sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.
For each test case, output $q$ integers: the $i$-th integer contains the optimal $r$ for the $i$-th query. If there are multiple solutions, output the smallest one.
Input
The first line of input contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The descriptions of the test cases follow.
The first line contains a single integer $n$ ($1 \le n \le 10^5$).
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^4$).
The third line contains a single integer $q$ ($1 \le q \le 10^5$).
The next $q$ lines each contain two integers $l$ and $u$ ($1 \le l \le n, 1 \le u \le 10^9$) — the descriptions to each query.
The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$. The sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.
Output
For each test case, output $q$ integers: the $i$-th integer contains the optimal $r$ for the $i$-th query. If there are multiple solutions, output the smallest one.
5
6
3 1 4 1 5 9
3
1 8
2 7
5 9
1
10
1
1 1
9
5 10 9 6 8 3 10 7 3
5
8 56
1 12
9 3
1 27
5 45
5
7 9 2 5 2
10
1 37
2 9
3 33
4 32
4 15
2 2
4 2
2 19
3 7
2 7
10
9 1 6 7 6 3 10 7 3 10
5
10 43
3 23
9 3
6 8
5 14
3 4 5
1
9 2 9 4 9
5 2 5 5 5 2 4 5 4 2
10 6 9 7 7
Note
For the $1$-st query in the first test case:
- By choosing $r = 3$, Isaac finishes $a_1 + a_2 + a_3 = 3 + 1 + 4 = 8$ sections in total, hence his increase in performance is $u+(u-1)+\ldots+(u-7)=8+7+6+5+4+3+2+1 = 36$.
- By choosing $r = 4$, Isaac finishes $a_1 + a_2 + a_3 + a_4 = 3 + 1 + 4 + 1 = 9$ sections in total, hence his increase in performance is $u+(u-1)+\ldots+(u-8)=8+7+6+5+4+3+2+1+0 = 36$.
Both choices yield the optimal increase in performance, however we want to choose the smallest $r$. So we choose $r = 3$.
For the $2$-nd query in the first test case, by choosing $r = 4$, Isaac finishes $a_2 + a_3 + a_4 = 1 + 4 + 1 = 6$ sections in total, hence his increase in performance is $u+(u-1)+\ldots+(u-5)=7+6+5+4+3+2 = 27$. This is the optimal increase in performance.
For the $3$-rd query in the first test case:
- By choosing $r = 5$, Isaac finishes $a_5 = 5$ sections in total, hence his increase in performance is $u+(u-1)+\ldots+(u-4)=9+8+7+6+5 = 35$.
- By choosing $r = 6$, Isaac finishes $a_5 + a_6 = 5 + 9 = 14$ sections in total, hence his increase in performance is $u+(u-1)+\ldots+(u-13)=9+8+7+6+5+4+3+2+1+0+(-1)+(-2)+(-3)+(-4) = 35$.
Both choices yield the optimal increase in performance, however we want to choose the smallest $r$. So we choose $r = 5$.
Hence the output for the first test case is $[3, 4, 5]$.