#include <bits/stdc++.h>
using namespace std;
const long double pi = 4 * atan((long double)1);
const long long mod = 1e9 + 7;
const long long inf = 922337203685477;
const long long nax = 5e5 + 5;
const long long nex = 1e6 + 5;
long long n, c, a[nax], pref[nax];
vector<vector<long long>> freq(nax);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> c;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == c) {
      pref[i]++;
    } else {
      freq[a[i]].push_back(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    pref[i] += pref[i - 1];
  }
  long long ans = pref[n];
  for (long long i = 1; i < nax; i++) {
    if (freq[i].size() > 0) {
      long long l = 0;
      for (long long r = 0; r < freq[i].size(); r++) {
        long long idxl = freq[i][l], idxr = freq[i][r];
        long long del = pref[idxr] - pref[idxl];
        long long plus = r - l + 1;
        while (del >= plus) {
          l++;
          idxl = freq[i][l];
          del = pref[idxr] - pref[idxl];
          plus = r - l + 1;
        }
        ans = max(ans, plus + pref[idxl - 1] + pref[n] - pref[idxr]);
      }
    }
  }
  cout << ans << '\n';
}

### 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;
const long double pi = 4 * atan((long double)1);
const long long mod = 1e9 + 7;
const long long inf = 922337203685477;
const long long nax = 5e5 + 5;
const long long nex = 1e6 + 5;
long long n, c, a[nax], pref[nax];
vector<vector<long long>> freq(nax);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> c;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == c) {
      pref[i]++;
    } else {
      freq[a[i]].push_back(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    pref[i] += pref[i - 1];
  }
  long long ans = pref[n];
  for (long long i = 1; i < nax; i++) {
    if (freq[i].size() > 0) {
      long long l = 0;
      for (long long r = 0; r < freq[i].size(); r++) {
        long long idxl = freq[i][l], idxr = freq[i][r];
        long long del = pref[idxr] - pref[idxl];
        long long plus = r - l + 1;
        while (del >= plus) {
          l++;
          idxl = freq[i][l];
          del = pref[idxr] - pref[idxl];
          plus = r - l + 1;
        }
        ans = max(ans, plus + pref[idxl - 1] + pref[n] - pref[idxr]);
      }
    }
  }
  cout << ans << '\n';
}

```