#include <bits/stdc++.h>
#pragma warning(disable : 6031)
#pragma warning(disable : 4101)
using namespace std;
const int INF = 1e9;
const long long LINF = 1e18;
const double PI = acos(-1);
const int MAXN = 300005;
int n, m, l, r, mid, A[MAXN];
int next_num(int cur, int num, int k) {
  if (num + k < m) {
    if (cur >= num && cur <= num + k) {
      return cur;
    } else if (cur > num + k) {
      return -1;
    } else {
      return num;
    }
  } else {
    if (cur <= (num + k) % m || cur >= num) {
      return cur;
    } else {
      return num;
    }
  }
}
bool is_possible(int k) {
  int start = A[0] + k < m ? A[0] : 0;
  for (int i = 1; i < n; i++) {
    start = next_num(start, A[i], k);
    if (start == -1) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
  }
  l = 0;
  r = m - 1;
  while (l != r) {
    mid = (l + r) / 2;
    if (is_possible(mid)) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  printf("%d\n", l);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

Toad Zitz has an array of integers, each integer is between 0 and m-1 inclusive. The integers are a_1, a_2, …, a_n.

In one operation Zitz can choose an integer k and k indices i_1, i_2, …, i_k such that 1 ≤ i_1 < i_2 < … < i_k ≤ n. He should then change a_{i_j} to ((a_{i_j}+1) mod m) for each chosen integer i_j. The integer m is fixed for all operations and indices.

Here x mod y denotes the remainder of the division of x by y.

Zitz wants to make his array non-decreasing with the minimum number of such operations. Find this minimum number of operations.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) — the number of integers in the array and the parameter m.

The next line contains n space-separated integers a_1, a_2, …, a_n (0 ≤ a_i < m) — the given array.

Output

Output one integer: the minimum number of described operations Zitz needs to make his array non-decreasing. If no operations required, print 0.

It is easy to see that with enough operations Zitz can always make his array non-decreasing.

Examples

Input


5 3
0 0 0 1 2


Output


0


Input


5 7
0 6 1 3 2


Output


1

Note

In the first example, the array is already non-decreasing, so the answer is 0.

In the second example, you can choose k=2, i_1 = 2, i_2 = 5, the array becomes [0,0,1,3,3]. It is non-decreasing, so the answer is 1.

### Response

```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 6031)
#pragma warning(disable : 4101)
using namespace std;
const int INF = 1e9;
const long long LINF = 1e18;
const double PI = acos(-1);
const int MAXN = 300005;
int n, m, l, r, mid, A[MAXN];
int next_num(int cur, int num, int k) {
  if (num + k < m) {
    if (cur >= num && cur <= num + k) {
      return cur;
    } else if (cur > num + k) {
      return -1;
    } else {
      return num;
    }
  } else {
    if (cur <= (num + k) % m || cur >= num) {
      return cur;
    } else {
      return num;
    }
  }
}
bool is_possible(int k) {
  int start = A[0] + k < m ? A[0] : 0;
  for (int i = 1; i < n; i++) {
    start = next_num(start, A[i], k);
    if (start == -1) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
  }
  l = 0;
  r = m - 1;
  while (l != r) {
    mid = (l + r) / 2;
    if (is_possible(mid)) {
      r = mid;
    } else {
      l = mid + 1;
    }
  }
  printf("%d\n", l);
  return 0;
}

```