831A- Unimodal Array

A. Unimodal Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Array of integers is unimodal, if:

• it is strictly increasing in the beginning; 
• after that it is constant; 
• after that it is strictly decreasing. 

The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.

For example, the following three arrays are unimodal: [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], but the following three are not unimodal: [5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].

Write a program that checks if an array is unimodal.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000) — the elements of the array.

Output

Print "YES" if the given array is unimodal. Otherwise, print "NO".

You can output each letter in any case (upper or lower).

Examples

input

6
1 5 5 5 4 2

output

YES

input

5
10 20 30 20 10

output

YES

input

4
1 2 1 2

output

NO

input

7
3 3 3 3 3 3 3

output

YES

Note

In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).

这一题读完之后我就顺着题意列出成立的5种情况，把数据做成线状图的话是：

1.先增后平

2.先增后减

3.先增后平后减

4.先平后减

5.一直平

然而这样实现的话太麻烦了，代码我都不好意思贴出来。

看了cf上高手的代码，一目了然。

#include <cstdio>
const int M = 1111;
int nu[M];
int main(){
    int n,f=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&nu[i]);
    }
    while(nu[f]<nu[f+1]&&f<n-1) f++;
    while(nu[f]==nu[f+1]&&f<n-1) f++;
    while(nu[f]>nu[f+1]&&f<n-1) f++;
    if(f==n-1) puts("YES");
    else puts("NO");
    return 0;
}