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).

解题说明：此题是一道模拟题，按照题目意思，序列保证先增后减即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iostream>
using namespace std;

int main()
{
int i, j, n;
int arr[102];
scanf("%d", &n);
for(i = 0; i < n; i++)
{
scanf("%d", &arr[i]);
}
i = 0;
j = 1;
while(j < n && arr[i] < arr[j])
{
i++;
j++;
}
while(j < n && arr[i] == arr[j])
{
i++;
j++;
}
while(j < n && arr[i] > arr[j])
{
i++;
j++;
}
if(j == n)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
return 0;
}