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.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 210000;
int n;
int a[maxn], b[maxn];
struct node {
  int ai, i;
} p[maxn];
inline bool cmp(const node &k1, const node &k2) {
  return k1.ai == k2.ai ? k1.i < k2.i : k1.ai < k2.ai;
}
int seg[maxn];
void build(const int x, const int l, const int r) {
  seg[x] = -1;
  if (l == r) return;
  int mid = (l + r) >> 1, lc = x << 1, rc = lc | 1;
  build(lc, l, mid);
  build(rc, mid + 1, r);
}
int loc, c;
void upd(const int x, const int l, const int r) {
  if (l == r) {
    seg[x] = max(seg[x], c);
    return;
  }
  int mid = (l + r) >> 1, lc = x << 1, rc = lc | 1;
  if (loc <= mid)
    upd(lc, l, mid);
  else
    upd(rc, mid + 1, r);
  seg[x] = max(seg[lc], seg[rc]);
}
int lx, rx;
int query(const int x, const int l, const int r) {
  if (rx < l || r < lx) return -1;
  if (lx <= l && r <= rx) return seg[x];
  int mid = (l + r) >> 1, lc = x << 1, rc = lc | 1;
  return max(query(lc, l, mid), query(rc, mid + 1, r));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[i].i = i, p[i].ai = a[i];
  }
  sort(p + 1, p + n + 1, cmp);
  build(1, 0, 2 * n);
  loc = n, c = 0, upd(1, 0, 2 * n);
  for (int i = 1; i <= n; i++) {
    lx = n + p[i].ai - p[i].i, rx = 2 * n;
    int k = query(1, 0, 2 * n);
    if (k != -1) {
      int kc = k + 1;
      loc = n + p[i].ai - p[i].i;
      c = kc;
      upd(1, 0, 2 * n);
    }
  }
  lx = 0, rx = 2 * n;
  cout << query(1, 0, 2 * n) << endl;
  return 0;
}
