Codeforces Round #424 (Div. 2) A. Unimodal Array（水题）

题目：

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).

思路：

判断一个数列是否单峰，先递增然后相等，然后递减，可能都相等

要注意最小的数不能有相等的,我就是因为这个WA了8次。。。。

代码：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 1000007
#define N 105
#define M 12357
#define ll long long
using namespace std;
int a
;
int n,maxn = 1;
bool check()
{
for(int i = 1; i <= n; i++)
{
if(a[i] == maxn)
{
for(int j = i-1; j >= 1; j--)
if(a[j] >= a[j+1])
return false;
while(a[i] == maxn && i <= n)
i++;
for(int j = i; j <= n; j++)
if(a[j-1] <= a[j] || a[j] >= maxn)
return false;
break;
}
}
return true;
}

int main()
{
cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i],maxn = max(a[i],maxn);
if(check())
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}