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 only horizontally on the neighbouring 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).

Note that Vova can't put bricks vertically.

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

NO

Input

2
10 10

Output

YES

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 no bricks in the wall.

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，磚只能橫著放

題解：相鄰兩個是相等高度的才能達到任意高度，像2112，若有未匹配成功的並且遇到了更高的也是不符合的，比如1221

最後若剩下一個未匹配成功的，那麼一定要是最高的，因為不能豎著放，高度是不可能增加的

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