Codeforces Round #424 (Div. 2) A. 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 e
ec4b
lements 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).

题意：判断一个数列是否为  增平减

分析：只要注意到几个点这题就是水题了

首先如果出现连续相等的数，那么后面不能再出现其它连续相等的数  1 1 2 2 1 1这样是不符合的。

AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
int a[2000],n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int i,f1=0,f2=0;
for(i=1;i<n;i++)
{
if(a[i]>a[i-1]&&!f2)
f1=1;
else if(a[i]==a[i-1])
f2=1;
else
break;
}
for(;i<n;i++)
{
if(a[i-1]<=a[i])
{
printf("NO\n");
return 0;
}
}
printf("YES\n");

}