#include <bits/stdc++.h>
using namespace std;
long long n, c, v[500001], f[500001], maxim;
vector<long long> poz[500001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> c;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    f[i] = f[i - 1];
    if (v[i] == c) f[i]++;
    poz[v[i]].push_back(i);
  }
  maxim = f[n];
  for (int d = 1; d <= 500000; d++)
    if (d != c && poz[d].size()) {
      long long summax = 1;
      long long suma = 1;
      for (int i = 1; i < poz[d].size(); i++) {
        long long p = poz[d][i];
        long long ant = poz[d][i - 1];
        suma -= (f[p - 1] - f[ant]);
        suma++;
        if (suma <= 0) suma = 1;
        summax = max(summax, suma);
      }
      if (f[n] + summax > maxim) maxim = f[n] + summax;
    }
  cout << maxim;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long n, c, v[500001], f[500001], maxim;
vector<long long> poz[500001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> c;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    f[i] = f[i - 1];
    if (v[i] == c) f[i]++;
    poz[v[i]].push_back(i);
  }
  maxim = f[n];
  for (int d = 1; d <= 500000; d++)
    if (d != c && poz[d].size()) {
      long long summax = 1;
      long long suma = 1;
      for (int i = 1; i < poz[d].size(); i++) {
        long long p = poz[d][i];
        long long ant = poz[d][i - 1];
        suma -= (f[p - 1] - f[ant]);
        suma++;
        if (suma <= 0) suma = 1;
        summax = max(summax, suma);
      }
      if (f[n] + summax > maxim) maxim = f[n] + summax;
    }
  cout << maxim;
  return 0;
}

```