Codeforces Round 431 (Div. 2)
A
Odds and Ends
Codeforces
1000
原题
←B →
翻译中…

Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?

翻译中…

Given an integer sequence $*a*_{1}, *a*_{2}, ..., *a*_{*n*}$ of length $*n*$. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.

翻译中…

A subsegment is a contiguous slice of the whole sequence. For example, ${3, 4, 5}$ and ${1}$ are subsegments of sequence ${1, 2, 3, 4, 5, 6}$, while ${1, 2, 4}$ and ${7}$ are not.

翻译中…

Input

翻译中…

The first line of input contains a non-negative integer $*n*$ ($1 ≤ *n* ≤ 100$) — the length of the sequence.

翻译中…

The second line contains $*n*$ space-separated non-negative integers $*a*_{1}, *a*_{2}, ..., *a*_{*n*}$ ($0 ≤ *a*_{*i*} ≤ 100$) — the elements of the sequence.

翻译中…

Output

翻译中…

Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.

翻译中…

You can output each letter in any case (upper or lower).

翻译中…

Note

翻译中…

In the first example, divide the sequence into $1$ subsegment: ${1, 3, 5}$ and the requirements will be met.

翻译中…

In the second example, divide the sequence into $3$ subsegments: ${1, 0, 1}$, ${5}$, ${1}$.

翻译中…

In the third example, one of the subsegments must start with $4$ which is an even number, thus the requirements cannot be met.

翻译中…

In the fourth example, the sequence can be divided into $2$ subsegments: ${3, 9, 9}$, ${3}$, but this is not a valid solution because $2$ is an even number.

翻译中…

样例

Input #1
3
1 3 5
Output #1
Yes
InputCopy OutputCopy
Input #2
5
1 0 1 5 1
Output #2
Yes
InputCopy OutputCopy
Input #3
3
4 3 1
Output #3
No
InputCopy OutputCopy
Input #4
4
3 9 9 3
Output #4
No
InputCopy OutputCopy
题目信息
题号
849A
时间限制
1 s
内存限制
256 MB
标签
模拟

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。