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 aa of nn integers, with aiai being the height of the ii-th part of the wall.

Vova can only use 2×12×1 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 ii the current height of part iiis the same as for part i+1i+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 11 of the wall or to the right of part nn 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 nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of parts in the wall.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤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

Copy

52 1 1 2 5

output

Copy

YES

input

Copy

34 5 3

output

Copy

YES

input

Copy

210 10

output

Copy

YES

input

Copy

31 2 3

output

Copy

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][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][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][4,5,5], then horizontally on parts 2 and 3 to make it [4,6,6][4,6,6] and then vertically on part 1 to make it [6,6,6][6,6,6].

In the third example the wall is already complete.

代码：

#include 
     
      using namespace std;int N;stack
      
        s;int main() {    scanf("%d", &N);    for(int i = 0; i < N; i ++) {        int x;        scanf("%d", &x);        if(s.empty())            s.push(x);        else {            if(abs(s.top() - x) % 2)                s.push(x);            else s.pop();        }    }    if(s.size() > 1) printf("NO\n");    else printf("YES\n");    return 0;}
      
     

　　相邻的两个墙的高度差是 2 的倍数就可以了