1">11 of the wall or to the right of part nn 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 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

5
2 1 1 2 5

output Copy

YES

input Copy

3
4 5 3

output Copy

NO

input Copy

2
10 10

output Copy

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][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 no bricks in the wall.

In the third example the wall is already complete.

 

題意概括：

給出 N 個牆的高度，只能用 2*1 的方塊去填，判斷是否能把所有的牆變成同一高度。

 

解題思路：

按順序遍歷，只有當兩個相鄰的牆高度相同時才能相互抵消。

細節就是如果出現有一個超長的牆出現阻隔，需要進行判斷這個超長的牆是否在端點，如果在端點則沒有影響，如果在中間則有影響。

 

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define FOR(x, maxx) for(i = 0; i < maxx; i++)
 9 using namespace std;
10 
11 const int MAXN = 2e5+10;
12 
13 LL stak[MAXN];
14 int top;
15 
16 int main()
17 {
18     int N, i;
19     bool flag = true;
20     LL x;
21     top = 0;
22     scanf("%d", &N);
23     scanf("%I64d", &x);
24     stak[++top] = x;
25     LL maa = x;
26     FOR(i, N-1){
27         scanf("%I64d", &x);
28         if(x > stak[top] && top > 0) flag = false;
29         else if(x == stak[top] && flag) top--;
30         else   stak[++top] = x;
31         maa = max(maa, x);
32     }
33 
34     if(top > 1 || !flag) puts("NO");
35     else{
36             if(top == 1 && stak[top] != maa) puts("NO");
37             else puts("YES");
38     }
39 
40     return 0;
41 }