There are n beautiful skyscrapers in New York, the height of the i-th one is h_i. Today some villains have set on fire first n - 1 of them, and now the only safety building is n-th skyscraper.

Let's call a jump from i-th skyscraper to j-th (i < j) discrete, if all skyscrapers between are strictly lower or higher than both of them. Formally, jump is discrete, if i < j and one of the following conditions satisfied: 

  * i + 1 = j 
  * max(h_{i + 1}, …, h_{j - 1}) < min(h_i, h_j) 
  * max(h_i, h_j) < min(h_{i + 1}, …, h_{j - 1}). 



At the moment, Vasya is staying on the first skyscraper and wants to live a little longer, so his goal is to reach n-th skyscraper with minimal count of discrete jumps. Help him with calcualting this number.

Input

The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — total amount of skyscrapers.

The second line contains n integers h_1, h_2, …, h_n (1 ≤ h_i ≤ 10^9) — heights of skyscrapers.

Output

Print single number k — minimal amount of discrete jumps. We can show that an answer always exists.

Examples

Input


5
1 3 1 4 5


Output


3

Input


4
4 2 2 4


Output


1

Input


2
1 1


Output


1

Input


5
100 1 100 1 100


Output


2

Note

In the first testcase, Vasya can jump in the following way: 1 → 2 → 4 → 5.

In the second and third testcases, we can reach last skyscraper in one jump.

Sequence of jumps in the fourth testcase: 1 → 3 → 5.
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const int MOD = 1e9 + 7;
const long double PI = 3.1415926535898;
const long long INF = LLONG_MAX;
vector<int> adj[(int)5e5 + 5];
int visited[(int)5e5 + 5] = {0};
int dis[(int)5e5 + 5];
void bfs() {
  dis[1] = 0;
  queue<int> q;
  q.push(1);
  visited[1] = 1;
  while (!q.empty()) {
    auto p = q.front();
    q.pop();
    for (auto to : adj[p]) {
      if (!visited[to]) {
        visited[to] = 1;
        dis[to] = dis[p] + 1;
        q.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  int h[n + 1];
  for (int i = 1; i <= n; i++) dis[i] = i - 1;
  for (int i = 1; i <= n; i++) cin >> h[i];
  stack<int> dec, inc;
  for (int i = 1; i <= n; i++) {
    int prev = 0;
    while (!dec.empty() && h[dec.top()] < h[i]) {
      if (prev != h[dec.top()]) adj[dec.top()].push_back(i);
      prev = h[dec.top()];
      dec.pop();
    }
    if (!dec.empty()) adj[dec.top()].push_back(i);
    dec.push(i);
    prev = 0;
    while (!inc.empty() && h[inc.top()] > h[i]) {
      if (prev != h[inc.top()]) adj[inc.top()].push_back(i);
      prev = h[inc.top()];
      inc.pop();
    }
    if (!inc.empty()) adj[inc.top()].push_back(i);
    inc.push(i);
  }
  bfs();
  cout << dis[n] << "\n";
}
