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.
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
int __builtin_popcount(int n) {
  int x = 0;
  while (n) n -= n & (-n), x++;
  return x;
}
using namespace std;
const int mod = 1000000009;
const int inf = 2012345678;
const long long infl = 9012345678901234567;
const double pi = 3.1415926535897932384626433832795;
const int N = 301010;
int a[N];
int main() {
  freopen("input.txt", "r", stdin);
  int n, m;
  scanf("%d%d", &(n), &(m));
  for (int i = 0; i < (n); i++) scanf("%d", &(a[i]));
  int low = 0, high = m;
  while (low != high) {
    int mid = low + high >> 1;
    bool good = true;
    int cur = 0;
    for (int i = 0; i < (n); i++) {
      if (a[i] + mid >= m) {
        if (cur >= a[i] || cur <= a[i] + mid - m)
          continue;
        else
          cur = a[i];
      } else {
        if (cur > a[i] + mid) {
          good = false;
          break;
        } else if (cur < a[i])
          cur = a[i];
      }
    }
    if (good)
      high = mid;
    else
      low = mid + 1;
  }
  printf("%d\n", low);
  return 0;
}
