There are n psychos standing in a line. Each psycho is assigned a unique integer from 1 to n. At each step every psycho who has an id greater than the psycho to his right (if exists) kills his right neighbor in the line. Note that a psycho might kill and get killed at the same step. 

You're given the initial arrangement of the psychos in the line. Calculate how many steps are needed to the moment of time such, that nobody kills his neighbor after that moment. Look notes to understand the statement more precise.

Input

The first line of input contains integer n denoting the number of psychos, (1 ≤ n ≤ 105). In the second line there will be a list of n space separated distinct integers each in range 1 to n, inclusive — ids of the psychos in the line from left to right.

Output

Print the number of steps, so that the line remains the same afterward.

Examples

Input

10
10 9 7 8 6 5 3 4 2 1


Output

2


Input

6
1 2 3 4 5 6


Output

0

Note

In the first sample line of the psychos transforms as follows: [10 9 7 8 6 5 3 4 2 1]  →  [10 8 4] →  [10]. So, there are two steps.
#include <bits/stdc++.h>
using namespace std;
int inf_int = 2e9;
long long inf_ll = 2e18;
const double pi = 3.1415926535898;
template <typename T, typename T1>
void prin(vector<pair<T, T1> >& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
template <typename T>
void prin(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i];
    if (i < a.size() - 1)
      cout << " ";
    else
      cout << "\n";
  }
}
template <typename T>
void prin(set<T>& a) {
  for (auto it = a.begin(); it != a.end(); it++) {
    cout << *it << " ";
  }
}
template <typename T>
void prin_new_line(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << "\n";
  }
}
template <typename T, typename T1>
void prin_new_line(vector<pair<T, T1> >& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
int sum_vec(vector<int>& a) {
  int s = 0;
  for (int i = 0; i < a.size(); i++) {
    s += a[i];
  }
  return s;
}
template <typename T>
T max(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = max(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = min(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <typename T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
long double s_triangle(long double x1, long double y1, long double x2,
                       long double y2, long double x3, long double y3) {
  return abs(((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)) / 2);
}
bool overflow(long long a, long long b) {
  if (a * b / b != a) return true;
  return false;
}
bool debug = 1;
const int maxn = 1001 * 1001;
long long mod = 1e9 + 7;
long double dis1[maxn] = {}, dis2[maxn] = {}, dis3[maxn] = {};
long double dis(long double gfgs, long double asd, long double x2,
                long double y2) {
  return sqrt((gfgs - x2) * (gfgs - x2) + (asd - y2) * (asd - y2));
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> t(n, 1);
  stack<int> s;
  s.push(0);
  t[0] = 0;
  int ans = 0;
  for (int i = 1; i < n; i++) {
    while (!s.empty() && a[s.top()] < a[i]) {
      t[i] = max(t[i], t[s.top()] + 1);
      s.pop();
    }
    if (s.empty()) t[i] = 0;
    s.push(i);
    ans = max(ans, t[i]);
  }
  cout << ans;
}
int main() {
  if (!debug) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  int t = 1;
  while (t--) solve();
  return 0;
}
