#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5;
long long mx[maxn], vis[maxn], pre[maxn], suf[maxn], a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, c;
  cin >> n >> c;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    suf[a[i]]++;
  }
  fill(mx, mx + maxn, -(1ll << 62));
  long long ans = suf[c] + (suf[c] != n);
  for (int i = 1; i <= n; i++) {
    pre[a[i]]++, suf[a[i]]--;
    if (a[i] != c && vis[a[i]]) {
      ans = max(ans, mx[a[i]] + pre[a[i]] + suf[c]);
    }
    if (a[i] != c) {
      vis[a[i]] = 1;
      mx[a[i]] = max(mx[a[i]], pre[c] - pre[a[i]] + 1);
    }
  }
  cout << ans << endl;
}

### Prompt

In CPP, your task is to solve 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 int maxn = 5e5 + 5;
long long mx[maxn], vis[maxn], pre[maxn], suf[maxn], a[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, c;
  cin >> n >> c;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    suf[a[i]]++;
  }
  fill(mx, mx + maxn, -(1ll << 62));
  long long ans = suf[c] + (suf[c] != n);
  for (int i = 1; i <= n; i++) {
    pre[a[i]]++, suf[a[i]]--;
    if (a[i] != c && vis[a[i]]) {
      ans = max(ans, mx[a[i]] + pre[a[i]] + suf[c]);
    }
    if (a[i] != c) {
      vis[a[i]] = 1;
      mx[a[i]] = max(mx[a[i]], pre[c] - pre[a[i]] + 1);
    }
  }
  cout << ans << endl;
}

```