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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <array>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <utility>
#include <functional>

using namespace std;
using lli = long long int;

int main() {
    long n;
    cin >> n;

    vector<long> p(n);
    for (long i = 0; i < n; ++i) {
        cin >> p[i];
    }

    int counter = 0;
    for (long i = 1; i < n - 1; ++i) {
        vector<long> check(3);
        check[0] = p[i - 1];
        check[1] = p[i];
        check[2] = p[i + 1];
        sort(check.begin(), check.end());
        if (check[1] == p[i]) {
            ++counter;
        }
    }

    cout << counter << endl;

    return 0;
}