Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches.

The current state of the wall can be respresented by a sequence a

of n integers, with ai being the height of the i

-th part of the wall.

Vova can only use 2×1

bricks to put in the wall (he has infinite supply of them, however).

Vova can put bricks horizontally on the neighboring parts of the wall of equal height. It means that if for some i

the current height of part i is the same as for part i+1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 1 of the wall or to the right of part n

of it).

The next paragraph is specific to the version 1 of the problem.

Vova can also put bricks vertically. That means increasing height of any part of the wall by 2.

Vova is a perfectionist, so he considers the wall completed when:

• all parts of the wall has the same height;
• the wall has no empty spaces inside it.

Can Vova complete the wall using any amount of bricks (possibly zero)?

Input

The first line contains a single integer n

(1≤n≤2⋅105

) — the number of parts in the wall.

The second line contains n

integers a1,a2,…,an (1≤ai≤109

) — the initial heights of the parts of the wall.

Output

Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero).

Print "NO" otherwise.

Examples

Input

5
2 1 1 2 5

Output

YES

Input

3
4 5 3

Output

YES

Input

2
10 10

Output

YES

Input

3
1 2 3

Output

NO

Note

In the first example Vova can put a brick on parts 2 and 3 to make the wall [2,2,2,2,5]

and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5,5,5,5,5]

.

In the second example Vova can put a brick vertically on part 3 to make the wall [4,5,5]

, then horizontally on parts 2 and 3 to make it [4,6,6] and then vertically on part 1 to make it [6,6,6]

.

In the third example the wall is already complete.

題意：給一排磚，每列的高度ai，問是否可以放1*2的磚，使得n列高度一樣，磚可以橫著放，也能豎著放

題解：相鄰的兩個的差值如果是偶數的話，可以先達到相等高度，然後就可以達到任意高度了，若兩邊也是差值為偶數，那麼就可以擴充套件到4個了，用棧維護一下，最後最多隻能有一個未匹配成功。

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int n,a[N];
int main()
{
	stack<int> s;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		if(!s.empty() && abs(s.top()-a[i])%2==0) s.pop();
		else s.push(a[i]);
	}
	if(s.size()<=1) printf("YES\n");
	else printf("NO\n");
	return 0;
}