#include <bits/stdc++.h>
const int maxn = 200005, K = 37, mod = 1e9 + 7;
int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
int dec(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
int mul(int a, int b) { return 1ll * a * b % mod; }
int ksm(int a, int b = mod - 2) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
int n, a[maxn];
std::tuple<int, int, int> p[maxn];
int f[maxn], c[maxn], mx, ans;
void ins(int x, int v) {
  for (; x <= mx; x += x & -x) c[x] = std::max(c[x], v);
}
int ask(int x) {
  int res = 0xcfcfcfcf;
  for (; x; x -= x & -x) res = std::max(res, c[x]);
  return res;
}
int main() {
  memset(c, 0xcf, sizeof(c));
  memset(f, 0xcf, sizeof(f));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", a + i), p[i] = std::make_tuple(i - a[i], a[i], i),
                        mx = std::max(mx, a[i]);
  mx++;
  std::sort(p + 1, p + 1 + n);
  ins(1, 0);
  for (int i = 1; i <= n; i++) {
    if (std::get<2>(p[i]) >= std::get<1>(p[i]))
      f[i] = ask(std::get<1>(p[i])) + 1;
    ins(std::get<1>(p[i]) + 1, f[i]);
    ans = std::max(ans, f[i]);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int maxn = 200005, K = 37, mod = 1e9 + 7;
int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
int dec(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
int mul(int a, int b) { return 1ll * a * b % mod; }
int ksm(int a, int b = mod - 2) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
int n, a[maxn];
std::tuple<int, int, int> p[maxn];
int f[maxn], c[maxn], mx, ans;
void ins(int x, int v) {
  for (; x <= mx; x += x & -x) c[x] = std::max(c[x], v);
}
int ask(int x) {
  int res = 0xcfcfcfcf;
  for (; x; x -= x & -x) res = std::max(res, c[x]);
  return res;
}
int main() {
  memset(c, 0xcf, sizeof(c));
  memset(f, 0xcf, sizeof(f));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", a + i), p[i] = std::make_tuple(i - a[i], a[i], i),
                        mx = std::max(mx, a[i]);
  mx++;
  std::sort(p + 1, p + 1 + n);
  ins(1, 0);
  for (int i = 1; i <= n; i++) {
    if (std::get<2>(p[i]) >= std::get<1>(p[i]))
      f[i] = ask(std::get<1>(p[i])) + 1;
    ins(std::get<1>(p[i]) + 1, f[i]);
    ans = std::max(ans, f[i]);
  }
  printf("%d\n", ans);
  return 0;
}

```