Codeforces Round 568 (Div. 2)
D
Extra Element
Codeforces
1700
原题
← C2E →
翻译中…

A sequence $a_1, a_2, \dots, a_k$ is called an arithmetic progression if for each $i$ from $1$ to $k$ elements satisfy the condition $a_i = a_1 + c \cdot (i - 1)$ for some fixed $c$.

翻译中…

For example, these five sequences are arithmetic progressions: $[5, 7, 9, 11]$, $[101]$, $[101, 100, 99]$, $[13, 97]$ and $[5, 5, 5, 5, 5]$. And these four sequences aren't arithmetic progressions: $[3, 1, 2]$, $[1, 2, 4, 8]$, $[1, -1, 1, -1]$ and $[1, 2, 3, 3, 3]$.

翻译中…

You are given a sequence of integers $b_1, b_2, \dots, b_n$. Find any index $j$ ($1 \le j \le n$), such that if you delete $b_j$ from the sequence, you can reorder the remaining $n-1$ elements, so that you will get an arithmetic progression. If there is no such index, output the number -1.

翻译中…

Input

翻译中…

The first line of the input contains one integer $n$ ($2 \le n \le 2\cdot10^5$) — length of the sequence $b$. The second line contains $n$ integers $b_1, b_2, \dots, b_n$ ($-10^9 \le b_i \le 10^9$) — elements of the sequence $b$.

翻译中…

Output

翻译中…

Print such index $j$ ($1 \le j \le n$), so that if you delete the $j$-th element from the sequence, you can reorder the remaining elements, so that you will get an arithmetic progression. If there are multiple solutions, you are allowed to print any of them. If there is no such index, print -1.

翻译中…

Note

翻译中…

Note to the first example. If you delete the $4$-th element, you can get the arithmetic progression $[2, 4, 6, 8]$.

翻译中…

Note to the second example. The original sequence is already arithmetic progression, so you can delete $1$-st or last element and you will get an arithmetical progression again.

翻译中…

样例

Input #1
5
2 6 8 7 4
Output #1
4
InputCopy OutputCopy
Input #2
8
1 2 3 4 5 6 7 8
Output #2
1
InputCopy OutputCopy
Input #3
4
1 2 4 8
Output #3
-1
InputCopy OutputCopy
题目信息
题号
1185D
时间限制
2 s
内存限制
256 MB
标签
模拟
数学

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