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.
题意：给一排砖，每列的高度ai，问是否可以放1*2的砖，使得n列高度一样，砖可以横着放，也能竖着放
题解：相邻的两个的差值如果是偶数的话，可以先达到相等高度，然后就可以达到任意高度了，若两边也是差值为偶数，那么就可以扩展到4个了，用栈维护一下，最后最多只能有一个未匹配成功。
#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;
}