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#include<bits/stdc++.h>
using namespace std;

#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define ST first
#define ND second
#define PB push_back
using ll = long long;
using ld = long double;
using pi = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;


#ifdef LOCAL
#define DTP(x, y)                                    \
    auto operator<<(auto &o, auto a)->decltype(y, o) \
    {                                                \
        o << "(";                                    \
        x;                                           \
        return o << ")";                             \
    }
DTP(o << a.first << ", " << a.second, a.second);
DTP(for (auto i : a) o << i << ", ", all(a));
void dump(auto... x) { ((cerr << x << ", "), ...) << '\n'; }
#define debug(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x)
#else
#define debug(...) 0
#endif



void Solve()
{
    int n;
    cin >> n;
    vi A(2*n);
    rep(i, 0, n)
        cin >> A[i];
    rep(i, n, 2*n-1)
        A[i]=A[i-n];
    A[2*n-1]=2000'000;
    int k=log2(n)+1;
    vvi P(2*n, vi(k+1));
    stack<pi> S;
    S.push(pi(A[2*n-1], 2*n-1));
    P[2*n-1][0]=2*n-1;
    for(int i=2*n-2; i>=0; --i){
        while(S.top().ST<=A[i])
            S.pop();
        P[i][0]=S.top().ND;
        S.push(pi(A[i], i));
    }
    rep(i, 1, k+1)
        rep(j, 0, 2*n)
            P[j][i]=P[P[j][i-1]][i-1];
    int res=1;
    rep(i, 0, n){
        int curr=1;
        int ind=i;
        for(int j=k; j>=0; --j){
            if(P[ind][j]<i+n){
                curr+=(1 << j);
                ind=P[ind][j];
            }
            if(curr+(1 << j)-1<res)
                break;
        }
        if(curr>res)
            res=curr;
    }
    cout << res << "\n";
    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    // cin >> t;
    while(t--)
    {
        Solve();
    }
    return 0;
}