You are given an array a consisting of n integers.

You can remove at most one element from this array. Thus, the final length of the array is n-1 or n.

Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.

Recall that the contiguous subarray a with indices from l to r is a[l ... r] = a_l, a_{l + 1}, ..., a_r. The subarray a[l ... r] is called strictly increasing if a_l < a_{l+1} < ... < a_r.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the maximum possible length of the strictly increasing contiguous subarray of the array a after removing at most one element.

Examples

Input


5
1 2 5 3 4


Output


4


Input


2
1 2


Output


2


Input


7
6 5 4 3 2 4 3


Output


2

Note

In the first example, you can delete a_3=5. Then the resulting array will be equal to [1, 2, 3, 4] and the length of its largest increasing subarray will be equal to 4.
import java.io.*;
import java.util.*;
 
public class utkarsh {
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);
    
    void solve() {
        int n = ni();
        int a[] = new int[n];
        for(int i = 0; i < n; i++)  a[i] = ni();
        int ans = 1;
        int dp[][] = new int[n][2];
        for(int i = 0; i < n; i++)  dp[i][0] = dp[i][1] = 1;
        for(int i = 1; i < n; i++) {
            if(a[i] > a[i-1]) {
                dp[i][0] = dp[i-1][0] + 1;
                dp[i][1] = dp[i-1][1] + 1;
            }
            if(i > 1 && a[i] > a[i-2]) {
                dp[i][1] = Math.max(dp[i][1], dp[i-2][0] + 1);
            }
            ans = Math.max(ans, Math.max(dp[i][0], dp[i][1]));
        }
        out.println(ans);
    }
    
    long mp(long b, long e) {
        long r = 1;
        while(e > 0) {
            if( (e&1) == 1 )    r = (r * b) % mod;
            b = (b * b) % mod;
            e >>= 1;
        }
        return r;
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }
    
    void run() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    public static void main(String[] args) {
        new utkarsh().run();
    }
}
