Problem:
Score : 200 points

Problem Statement

We have a permutation p = {p_1,\ p_2,\ ...,\ p_n} of {1,\ 2,\ ...,\ n}.

Print the number of elements p_i (1 < i < n) that satisfy the following condition:

p_i is the second smallest number among the three numbers p_{i - 1}, p_i, and p_{i + 1}.

Constraints

All values in input are integers.

3 \leq n \leq 20

p is a permutation of {1,\ 2,\ ...,\ n}.

Input

Input is given from Standard Input in the following format:

n
p_1 p_2 ... p_n

Output

Print the number of elements p_i (1 < i < n) that satisfy the condition.

Sample Input 1

5
1 3 5 4 2

Sample Output 1

2

p_2 = 3 is the second smallest number among p_1 = 1, p_2 = 3, and p_3 = 5. Also, p_4 = 4 is the second smallest number among p_3 = 5, p_4 = 4, and p_5 = 2. These two elements satisfy the condition.

Sample Input 2

9
9 6 3 2 5 8 7 4 1

Sample Output 2

5

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#define loop(i, l, r) for(i=l; i<r; i++)/*Instead of function, (for).*/
#define rep(i,r) for(i=0; i<r; i++)
#define Min(a, b) ((a < b)? a : b)
#define Max(a, b) ((a > b)? a : b)
#define Abs(x) ((x>=0)? x : -x)/*Return the absolute value of x.*/
typedef long long int lli;
/*Return Combination ,(nCr)*/
int combi(int n, int r)
{
	if(r==0 || r==n) return 1;
	else return combi(n-1,r) + combi(n-1,r-1);
}
/*Return gcd(a, b)*/
int gcd(int a, int b){
	if(b == 0) return a;
	else return gcd(b, a % b);
}
int main(void)
{
	int n, i, k=0;
	scanf("%d", &n);
	int p[n];
	rep(i,n) scanf("%d", &p[i]);
	loop(i, 1, n-1){
		if((p[i-1] < p[i]) &&(p[i] < p[i+1]) || (p[i+1] < p[i]) && (p[i] < p[i-1])){
			k++;
		}
	}
	printf("%d", k);
	return 0;
}