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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//#include <tr1/unordered_map>
//#include"Bignum/bignum.h"
//#define big bignum
#define lowbit(x) (x & -x)
#define debug(x) (cout << "#x = " << (x) << endl)
#define Set(x, i) memset (x, i, sizeof(x))
#define R register
#define For(i, j, k) for(R int i = (j), ED = (k); i <= ED; ++i)
#define Rep(i, j, k) for(R int i = (j), ED = (k); i >= ED; --i)
#define Cross(i, j, k) for(R int i = (j); i; i = (k))
using namespace std;
typedef long long ll;
const ll N = 100011;
const ll INF = 5e16;

namespace IO {

	inline char gc() {
		static char buf[100000], *p1 = buf, *p2 = buf;
		return (p1 == p2) && (p2 = (p1 = buf) +
			fread (buf, 1, 100000, stdin), p1 == p2)? EOF: *p1++;
	}

	#define dd ch = getchar()
	inline ll read() {
		R ll x = 0; R int f = 0; R char dd;
		for (; !isdigit (ch); dd) f ^= (ch == '-');
		for (; isdigit (ch); dd)  x = x * 10 + (ch ^ 48);
		return f? -x: x;
	}
	#undef dd

	inline void write( ll x ) {
		if (x < 0) putchar ('-'), x = -x;
		if (x > 9) write (x / 10); putchar (x % 10 | 48);
	}

	inline void wrn ( ll x ) { write (x); putchar (' '); }

	inline void wln ( ll x ) { write (x); putchar ('\n'); }

	inline void wlnn ( ll x, ll y ) { wrn (x), wln (y); }

}

using IO::wln;
using IO::read;

namespace Cesare {
	
	ll Ans = 0;
	ll n, p[N];
	
	int main() {
		n = read();
		For ( i, 1, n ) p[i] = read();
		For ( i, 2, n - 1 ) 
			if (p[i] > p[i - 1] && p[i] < p[i + 1] || 
				p[i] < p[i - 1] && p[i] > p[i + 1]) ++Ans;
		return wln (Ans), 0;
	}
	
}

int main()
{
	return Cesare::main();
}

/*

*/