#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, c, tot;
int a[500050], t[500050];
unordered_map<int, vector<int>> f;
int num[500050], ans;
int main() {
  cin >> n >> c;
  for (int i = 1; i <= n; i++) a[i] = read(), a[i] = c - a[i], num[i] = a[i];
  sort(num + 1, num + 1 + n);
  for (int i = 1; i <= n; i++)
    if ((i == 1 || num[i] != num[i - 1]) && num[i]) num[++tot] = num[i];
  for (int i = 1; i <= n; i++) {
    t[i] = t[i - 1];
    if (a[i] == 0)
      t[i]++;
    else
      f[a[i]].push_back(i);
  }
  ans = t[n];
  for (int i = 1; i <= tot; i++) {
    int mx = 0, mn = 1000000000;
    for (int j = 0; j < f[num[i]].size(); j++) {
      int val = j + 1 - t[f[num[i]][j]];
      mx = max(max(val - mn + 1, 1), mx);
      mn = min(mn, val);
    }
    ans = max(ans, mx + t[n]);
  }
  printf("%d\n", ans);
}

### Prompt

Generate a Cpp solution to the following problem:

You are given array a of length n. You can choose one segment [l, r] (1 ≤ l ≤ r ≤ n) and integer value k (positive, negative or even zero) and change a_l, a_{l + 1}, ..., a_r by k each (i.e. a_i := a_i + k for each l ≤ i ≤ r).

What is the maximum possible number of elements with value c that can be obtained after one such operation?

Input

The first line contains two integers n and c (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ c ≤ 5 ⋅ 10^5) — the length of array and the value c to obtain.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5 ⋅ 10^5) — array a.

Output

Print one integer — the maximum possible number of elements with value c which can be obtained after performing operation described above.

Examples

Input


6 9
9 9 9 9 9 9


Output


6


Input


3 2
6 2 6


Output


2

Note

In the first example we can choose any segment and k = 0. The array will stay same.

In the second example we can choose segment [1, 3] and k = -4. The array will become [2, -2, 2].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, c, tot;
int a[500050], t[500050];
unordered_map<int, vector<int>> f;
int num[500050], ans;
int main() {
  cin >> n >> c;
  for (int i = 1; i <= n; i++) a[i] = read(), a[i] = c - a[i], num[i] = a[i];
  sort(num + 1, num + 1 + n);
  for (int i = 1; i <= n; i++)
    if ((i == 1 || num[i] != num[i - 1]) && num[i]) num[++tot] = num[i];
  for (int i = 1; i <= n; i++) {
    t[i] = t[i - 1];
    if (a[i] == 0)
      t[i]++;
    else
      f[a[i]].push_back(i);
  }
  ans = t[n];
  for (int i = 1; i <= tot; i++) {
    int mx = 0, mn = 1000000000;
    for (int j = 0; j < f[num[i]].size(); j++) {
      int val = j + 1 - t[f[num[i]][j]];
      mx = max(max(val - mn + 1, 1), mx);
      mn = min(mn, val);
    }
    ans = max(ans, mx + t[n]);
  }
  printf("%d\n", ans);
}

```