#include <bits/stdc++.h>
using namespace std;
namespace io {
char ibuf[(1 << 20)], *iS, *iT, obuf[(1 << 20)],
    *oS = obuf, *oT = oS + (1 << 20) - 1, c, qu[55];
int f, qr;
inline void flush(void) {
  return fwrite(obuf, 1, oS - obuf, stdout), oS = obuf, void();
}
inline char getch(void) {
  return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20), stdin),
                      (iS == iT ? EOF : *iS++))
                   : *iS++);
}
inline void putch(char x) {
  *oS++ = x;
  if (oS == oT) flush();
  return;
}
string getstr(void) {
  string s = "";
  char c = getch();
  while (c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF)
    c = getch();
  while (!(c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF))
    s.push_back(c), c = getch();
  return s;
}
void putstr(string str, int begin = 0, int end = -1) {
  if (end == -1) end = str.size();
  for (register int i = begin; i < end; i++) putch(str[i]);
  return;
}
template <typename T>
inline T read() {
  register T x = 0;
  for (f = 1, c = getch(); c < '0' || c > '9'; c = getch())
    if (c == '-') f = -1;
  for (x = 0; c <= '9' && c >= '0'; c = getch()) x = x * 10 + (c & 15);
  return x * f;
}
template <typename T>
inline void write(const T& t) {
  register T x = t;
  if (!x) putch('0');
  if (x < 0) putch('-'), x = -x;
  while (x) qu[++qr] = x % 10 + '0', x /= 10;
  while (qr) putch(qu[qr--]);
  return;
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher_;
}  // namespace io
using io::getch;
using io::getstr;
using io::putch;
using io::putstr;
using io::read;
using io::write;
bool mem1;
pair<int, int> a[200005];
int S[200005];
void solve(void) {
  int n = read<int>();
  for (int i = 1; i <= n; i++) {
    int x = read<int>();
    a[i] = {i - x, x};
  }
  sort(a + 1, a + n + 1);
  int answer = 0;
  memset(S, 0x3f, sizeof(S));
  for (int i = 1; i <= n; i++) {
    if (a[i].first < 0) continue;
    int x = a[i].second, p = lower_bound(S + 1, S + answer + 1, x) - S;
    S[p] = min(S[p], x);
    answer = max(answer, p);
  }
  write(answer), putch('\n');
  return;
}
bool mem2;
int main() {
  clock_t t1 = clock();
  cerr << "Memory: " << abs(&mem1 - &mem2) / 1024.0 / 1024.0 << "MB" << endl;
  int _ = 1;
  while (_--) solve();
  clock_t t2 = clock();
  cerr << "Time: " << (t2 - t1) * 1000.0 / CLOCKS_PER_SEC << "ms" << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
namespace io {
char ibuf[(1 << 20)], *iS, *iT, obuf[(1 << 20)],
    *oS = obuf, *oT = oS + (1 << 20) - 1, c, qu[55];
int f, qr;
inline void flush(void) {
  return fwrite(obuf, 1, oS - obuf, stdout), oS = obuf, void();
}
inline char getch(void) {
  return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20), stdin),
                      (iS == iT ? EOF : *iS++))
                   : *iS++);
}
inline void putch(char x) {
  *oS++ = x;
  if (oS == oT) flush();
  return;
}
string getstr(void) {
  string s = "";
  char c = getch();
  while (c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF)
    c = getch();
  while (!(c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF))
    s.push_back(c), c = getch();
  return s;
}
void putstr(string str, int begin = 0, int end = -1) {
  if (end == -1) end = str.size();
  for (register int i = begin; i < end; i++) putch(str[i]);
  return;
}
template <typename T>
inline T read() {
  register T x = 0;
  for (f = 1, c = getch(); c < '0' || c > '9'; c = getch())
    if (c == '-') f = -1;
  for (x = 0; c <= '9' && c >= '0'; c = getch()) x = x * 10 + (c & 15);
  return x * f;
}
template <typename T>
inline void write(const T& t) {
  register T x = t;
  if (!x) putch('0');
  if (x < 0) putch('-'), x = -x;
  while (x) qu[++qr] = x % 10 + '0', x /= 10;
  while (qr) putch(qu[qr--]);
  return;
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher_;
}  // namespace io
using io::getch;
using io::getstr;
using io::putch;
using io::putstr;
using io::read;
using io::write;
bool mem1;
pair<int, int> a[200005];
int S[200005];
void solve(void) {
  int n = read<int>();
  for (int i = 1; i <= n; i++) {
    int x = read<int>();
    a[i] = {i - x, x};
  }
  sort(a + 1, a + n + 1);
  int answer = 0;
  memset(S, 0x3f, sizeof(S));
  for (int i = 1; i <= n; i++) {
    if (a[i].first < 0) continue;
    int x = a[i].second, p = lower_bound(S + 1, S + answer + 1, x) - S;
    S[p] = min(S[p], x);
    answer = max(answer, p);
  }
  write(answer), putch('\n');
  return;
}
bool mem2;
int main() {
  clock_t t1 = clock();
  cerr << "Memory: " << abs(&mem1 - &mem2) / 1024.0 / 1024.0 << "MB" << endl;
  int _ = 1;
  while (_--) solve();
  clock_t t2 = clock();
  cerr << "Time: " << (t2 - t1) * 1000.0 / CLOCKS_PER_SEC << "ms" << endl;
  return 0;
}

```