P

P - Psychos in a Line

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Appoint description:  System Crawler  (2013-07-04)

Description

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 nspace 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.

Sample Input

Input

10
10 9 7 8 6 5 3 4 2 1

Output

2

Input

6
1 2 3 4 5 6

Output

0

 

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 301000;
typedef pair<int,int> ii;
int a[maxn];
int next[maxn];//链表指针
int main() 
{
    //freopen("in.txt","r",stdin);
    int n;
    while(cin>>n)
    {
        queue<ii> q;
        rep(i,0,n)
        {
            scanf("%d",&a[i]);
            next[i] = i+1;
        }
        
        repd(i,n-2,0)
        {
            if(a[i] > a[i+1])
                q.push(ii(i,0));
        }
        
        int ans = 0;
        while(!q.empty())
        {
            int pos = q.front().first;
            int cas = q.front().second;
            ans = max(ans,cas);
            q.pop();
            if(next[pos] != n && a[pos] > a[next[pos]])
            {
                q.push(ii(pos,cas+1));
                next[pos] = next[next[pos]];
            }
        }
        
        cout<<ans<<endl;
    }
    return 0;
}