#include <bits/stdc++.h>
using namespace std;
vector<int> e[500005];
int a[500005], p[500005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    e[a[i]].push_back(i);
    p[i] = p[i - 1] + (a[i] == k);
  }
  int ans = 0;
  for (int z = 1; z <= 500000; ++z) {
    if (!e[z].size() || z == k) continue;
    int l = 0;
    for (int r = 0; r < e[z].size(); ++r) {
      while ((r - l + 1 - (p[e[z][r]] - p[e[z][l]])) <= 0) ++l;
      ans = max(ans, r - l + 1 + p[n] - (p[e[z][r]] - p[e[z][l]]));
    }
  }
  cout << max(ans, p[n]);
  return 0;
}

### Prompt

Please formulate 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;
vector<int> e[500005];
int a[500005], p[500005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    e[a[i]].push_back(i);
    p[i] = p[i - 1] + (a[i] == k);
  }
  int ans = 0;
  for (int z = 1; z <= 500000; ++z) {
    if (!e[z].size() || z == k) continue;
    int l = 0;
    for (int r = 0; r < e[z].size(); ++r) {
      while ((r - l + 1 - (p[e[z][r]] - p[e[z][l]])) <= 0) ++l;
      ans = max(ans, r - l + 1 + p[n] - (p[e[z][r]] - p[e[z][l]]));
    }
  }
  cout << max(ans, p[n]);
  return 0;
}

```