Mr. Chanek gives you a sequence a indexed from 1 to n. Define f(a) as the number of indices where a_i = i. 

You can pick an element from the current sequence and remove it, then concatenate the remaining elements together. For example, if you remove the 3-rd element from the sequence [4, 2, 3, 1], the resulting sequence will be [4, 2, 1]. 

You want to remove some elements from a in order to maximize f(a), using zero or more operations. Find the largest possible f(a).

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the initial length of the sequence.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the initial sequence a.

Output

Output an integer denoting the largest f(a) that can be obtained by doing zero or more operations.

Examples

Input


7
2 1 4 2 5 3 7


Output


3


Input


4
4 2 3 1


Output


2

Note

In the first example, f(A) = 3 by doing the following operations.

[2,1,4,2,5,3,7] → [2,1,2,5,3,7] → [1,2,5,3,7] → [1,2,5,3] → [1,2,3]

In the second example, f(A) = 2 and no additional operation is needed.

#include <bits/stdc++.h>  
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;   
using namespace std;
typedef long long ll;
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> 


class comp{
    public:
    bool operator()(const pair<ll,ll> &a, pair<ll,ll> &b){
        if(a.first <= b.first && a.second <= b.second)
            return true;
        return false;
    }
};

int lis(vector<pair<ll,ll>> const& a) {
    int n = a.size();
    const ll INF = 1e9;
    vector<pair<ll,ll>> d(n+1, {INF,INF});
    d[0] = {-INF,-INF};

    for (int i = 0; i < n; i++) {
        int j = upper_bound(d.begin(), d.end(), a[i],comp()) - d.begin();
        if (d[j-1].second <= a[i].second && d[j-1].first <= a[i].first && a[i].first <= d[j].first && a[i].second <= d[j].second)
            d[j] = a[i];
    }

    int ans = 0;
    for (int i = 0; i <= n; i++) {
        if(d[i] < make_pair(INF,INF))
            ans = i;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll n;
    cin>>n;
    vector<ll> arr(n);
    for(ll i =0;i<n;i++)
        cin>>arr[i];
    
    vector<pair<ll,ll>> arr2;
    for(ll i = 0;i<n;i++){
        ll val = i+1 - arr[i];
        if(val >=0)
        arr2.push_back({val,arr[i]});
    }
    //for(auto i:arr)
    //    cout<<i<<" ";
    //cout<<endl;
    //for(auto i:arr2)
    //    cout<<i.first<<i.second<<" ";
    //cout<<endl;
    
    cout<<lis(arr2)<<endl;
    return 0;
}