DZY has a sequence a, consisting of n integers.

We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In a single line print the answer to the problem — the maximum length of the required subsegment.

Examples

Input

6
7 2 3 1 5 6


Output

5

Note

You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4.
#include <bits/stdc++.h>
using namespace std;
long long int Mmax(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int Mmin(long long int a, long long int b) {
  if (a < b) return a;
  return b;
}
long long int nod(long long int a, long long int b) {
  while (a && b) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return Mmax(a, b);
}
long long int nok(long long int a, long long int b) {
  return a * b / nod(a, b);
}
bool IsPrime(long long int x) {
  if (x < 2) return false;
  long long int X = sqrt(x), i;
  for (i = 2; i <= X; i++)
    if (x % i == 0) return false;
  return true;
}
void hanoi(int n, int A, int C, int B) {
  if (n == 1) {
    cout << n << " " << A << " " << C << endl;
  } else {
    hanoi(n - 1, A, B, C);
    cout << n << " " << A << " " << C << endl;
    hanoi(n - 1, B, C, A);
  }
}
string pr2(string a, int d) {
  if (d == 0) return "0";
  string b;
  long long int sz = a.size(), i, prenos = 0;
  for (i = 0; i < sz; i++) {
    b[i] = a[i];
  }
  for (i = sz - 1; i > -1; i--) {
    a[i] = ((b[i] - '0') * d + prenos) % 10 + '0';
    prenos = ((b[i] - '0') * d + prenos) / 10;
  }
  if (prenos) a = char(prenos + '0') + a;
  return a;
}
string sum(string a, string b) {
  bool carry = false;
  long long int i, sz1, sz2, maxsz, minsz;
  string c, d;
  sz1 = a.size();
  sz2 = b.size();
  maxsz = max(sz1, sz2);
  minsz = min(sz1, sz2);
  while (sz1 < maxsz) {
    sz1++;
    a = '0' + a;
  }
  while (sz2 < maxsz) {
    sz2++;
    b = '0' + b;
  }
  for (i = maxsz - 1; i > -1; i--) {
    d = char((a[i] + b[i] - 96 + carry) % 10 + 48) + d;
    if (a[i] + b[i] - 96 + carry > 9)
      carry = true;
    else
      carry = false;
  }
  if (carry == true) d = char('1') + d;
  return d;
}
string pr(string a, string b) {
  string res = "0", p, p2;
  int sz = a.size(), x = 0;
  for (sz = a.size(); sz > 0; sz--, x++) {
    int d = a[sz - 1] - '0';
    a = a.erase(sz - 1, 1);
    p2 = pr2(b, d);
    p2 += p;
    res = sum(res, p2);
    p += "0";
  }
  return res;
}
bool vesokosna(long long int x) {
  return (x % 4 == 0 && x % 100 || x % 400 == 0);
}
long long int reverse(long long int x) {
  long long int mirror = 0;
  while (x) {
    mirror = mirror * 10 + x % 10;
    x /= 10;
  }
  return mirror;
}
long long int ost(string x, long long int k) {
  long long int num = 0, i, sz = x.size();
  for (i = 0; i < sz; i++) {
    num = num * 10;
    num += x[i] - '0';
    num %= k;
  }
  return num;
}
int a[131072];
int p[131072];
int p2[131072];
int main() {
  int n, i, ans = 0, here;
  cin >> n;
  p[0] = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > a[i - 1])
      p[i] = p[i - 1] + 1;
    else
      p[i] = 1;
    p2[i] = p[i];
    if (p[i] > ans) {
      ans = p[i];
      here = 1;
    }
  }
  for (i = n; i >= 1; i--) {
    if (a[i] < a[i + 1]) p[i] = p[i + 1];
  }
  for (i = n - 1; i > 0; i--) {
    if (a[i + 1] - a[i - 1] > 1 && a[i - 1] >= a[i] ||
        a[i] >= a[i + 1] && a[i + 1] > a[i - 1] + 1) {
      if (p[i + 1] + p[i - 1] > ans) {
        here = 2;
        ans = p[i + 1] + p[i - 1];
      }
    }
  }
  if (here == 1 && ans == n)
    cout << ans << endl;
  else if (here == 1)
    cout << ans + 1 << endl;
  else
    cout << ans << endl;
  return 0;
}
