#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const long long INF = 1000000000;
template <class T>
T read(T &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= flag;
  return x;
}
map<int, int> freq, mn;
int a[500005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, c;
  cin >> n >> c;
  int mx = 0, mn2 = 0;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    int diff = a[i + 1] - c;
    if (diff != 0) {
      mn[diff] = min(mn[diff], freq[diff] - freq[0]);
    }
    freq[diff]++;
    int gm = (freq[diff] - freq[0]) - mn[diff];
    mx = max(mx, gm);
  }
  cout << max(freq[0], freq[0] + mx) << endl;
}

### Prompt

Your challenge is to write 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 long MOD = 1000000007;
const long long INF = 1000000000;
template <class T>
T read(T &x) {
  char c;
  int flag = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag *= -1;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= flag;
  return x;
}
map<int, int> freq, mn;
int a[500005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, c;
  cin >> n >> c;
  int mx = 0, mn2 = 0;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    int diff = a[i + 1] - c;
    if (diff != 0) {
      mn[diff] = min(mn[diff], freq[diff] - freq[0]);
    }
    freq[diff]++;
    int gm = (freq[diff] - freq[0]) - mn[diff];
    mx = max(mx, gm);
  }
  cout << max(freq[0], freq[0] + mx) << endl;
}

```