Codeforces

Odds and Ends

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Given an integer sequence a1, a2, ..., an 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.

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.

Input

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

The second line contains n space-separated non-negative integers a1, a2, ..., an (0 ≤ ai ≤ 100) — the elements of the sequence.

Output

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

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

Examples

input

31 3 5

output

Yes

input

51 0 1 5 1

output

Yes

input

34 3 1

output

No

input

43 9 9 3

output

No

Note

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

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

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

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.

题意：给你一堆数，问你能不能分割成奇数个个数为奇数的子串，且所有子串的头尾都为奇数。

解题思路：乍一看好像很难……一开始我用搜索，枚举区间去做，妥妥的被hack了，超时了。实际上稍微思考就会发现，只要判断几个简单的条件就可以了。

奇数个奇数相加肯定是奇数（偶数不可能分解成奇数个奇数相加）（不考虑负数）

基于这一点，只要n是偶数，那么肯定输出No

那么是奇数的话……直接看头尾是不是奇数就可以了……因为头尾其中一个不是奇数的话，那么肯定输出No……

#include<iostream>#include<deque>#include<memory.h>#include<stdio.h>#include<map>#include<string>#include<algorithm>#include<vector>#include<math.h>#include<stack>#include<queue>#include<set>#define INF 1<<29using namespace std;int main(){    int n;    cin >> n;    int a[n+2];    for(int i=0;i<n;i++)        cin >> a[i];    if(n%2 && a[0]%2 && a[n-1]%2)        cout << "Yes" << endl;    else        cout << "No" << endl;}