#include <bits/stdc++.h>
using namespace std;
const int mxN = 2e5 + 5;
vector<int> a[mxN];
int tree[4 * mxN];
void update(int x, int lx, int rx, int i, int v) {
  if (lx == rx) {
    tree[x] = max(tree[x], v);
    return;
  }
  int mx = (lx + rx) >> 1;
  if (i <= mx)
    update(x << 1, lx, mx, i, v);
  else
    update(x << 1 | 1, mx + 1, rx, i, v);
  tree[x] = max(tree[x << 1], tree[x << 1 | 1]);
}
int query(int x, int lx, int rx, int l, int r) {
  if (l > rx || lx > r) return 0;
  if (l <= lx && r >= rx) return tree[x];
  int mx = (lx + rx) >> 1;
  return max(query(x << 1, lx, mx, l, r), query(x << 1 | 1, mx + 1, rx, l, r));
}
void testCase() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    reverse(a[i].begin(), a[i].end());
    vector<int> ans(a[i].size());
    for (int j = 0; j < a[i].size(); j++) {
      if (i - a[i][j] > 0) continue;
      ans[j] = query(1, 1, n, 1, a[i][j] - i + 1) + 1;
    }
    for (int j = 0; j < a[i].size(); j++) {
      if (i - a[i][j] > 0) continue;
      update(1, 1, n, a[i][j] - i + 1, ans[j]);
    }
  }
  cout << query(1, 1, n, 1, n) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  testCase();
  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>
using namespace std;
const int mxN = 2e5 + 5;
vector<int> a[mxN];
int tree[4 * mxN];
void update(int x, int lx, int rx, int i, int v) {
  if (lx == rx) {
    tree[x] = max(tree[x], v);
    return;
  }
  int mx = (lx + rx) >> 1;
  if (i <= mx)
    update(x << 1, lx, mx, i, v);
  else
    update(x << 1 | 1, mx + 1, rx, i, v);
  tree[x] = max(tree[x << 1], tree[x << 1 | 1]);
}
int query(int x, int lx, int rx, int l, int r) {
  if (l > rx || lx > r) return 0;
  if (l <= lx && r >= rx) return tree[x];
  int mx = (lx + rx) >> 1;
  return max(query(x << 1, lx, mx, l, r), query(x << 1 | 1, mx + 1, rx, l, r));
}
void testCase() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    reverse(a[i].begin(), a[i].end());
    vector<int> ans(a[i].size());
    for (int j = 0; j < a[i].size(); j++) {
      if (i - a[i][j] > 0) continue;
      ans[j] = query(1, 1, n, 1, a[i][j] - i + 1) + 1;
    }
    for (int j = 0; j < a[i].size(); j++) {
      if (i - a[i][j] > 0) continue;
      update(1, 1, n, a[i][j] - i + 1, ans[j]);
    }
  }
  cout << query(1, 1, n, 1, n) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  testCase();
  return 0;
}

```