Problem:
Score : 700 points

Problem Statement

Snuke is buying a lamp.
The light of the lamp can be adjusted to m levels of brightness, represented by integers from 1 through m, by the two buttons on the remote control.

The first button is a "forward" button. When this button is pressed, the brightness level is increased by 1, except when the brightness level is m, in which case the brightness level becomes 1.

The second button is a "favorite" button. When this button is pressed, the brightness level becomes the favorite brightness level x, which is set when the lamp is purchased.

Snuke is thinking of setting the favorite brightness level x so that he can efficiently adjust the brightness.
He is planning to change the brightness n-1 times. In the i-th change, the brightness level is changed from a_i to a_{i+1}. The initial brightness level is a_1.
Find the number of times Snuke needs to press the buttons when x is set to minimize this number.

Constraints

2 \leq n,m \leq 10^5

1 \leq a_i\leq m

a_i \neq a_{i+1}

n, m and a_i are integers.

Input

Input is given from Standard Input in the following format:

n m
a_1 a_2 … a_n

Output

Print the minimum number of times Snuke needs to press the buttons.

Sample Input 1

4 6
1 5 1 4

Sample Output 1

5

When the favorite brightness level is set to 1, 2, 3, 4, 5 and 6, Snuke needs to press the buttons 8, 9, 7, 5, 6 and 9 times, respectively.
Thus, Snuke should set the favorite brightness level to 4.
In this case, the brightness is adjusted as follows:

In the first change, press the favorite button once, then press the forward button once.

In the second change, press the forward button twice.

In the third change, press the favorite button once.

Sample Input 2

10 10
10 9 8 7 6 5 4 3 2 1

Sample Output 2

45

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
// #define int long long
// #define double long double
#define rep(i,n) for (int i=0; i<(int)(n); ++i)
#define rep1(i,n) for (int i=1; i<(int)(n); ++i)
#define repeq(i,n) for (int i=0; i<=(int)(n); ++i)
#define rep1eq(i,n) for (int i=1; i<=(int)(n); ++i)
#define rrep(i,n) for (int i=(int)(n)-1; i>=0; --i)
#define rrep1(i,n) for (int i=(int)(n)-1; i>0; --i)
#define rrepeq(i,n) for (int i=(int)(n); i>=0; --i)
#define rrep1eq(i,n) for (int i=(int)(n); i>0; --i)
#define REP(i,a,b) for (int i=(int)(a); i<=(int)(b); ++i)
#define RREP(i,a,b) for (int i=(int)(a); i>=(int)(b); --i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
template<typename T> using Graph = vector<vector<T>>;
template<typename T> using Spacial = vector<vector<vector<T>>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1e9+7;
const int MOD2 = 998244353;
// const double EPS = 1e-9;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string interval[2] = {" ", "\n"}; // {" ", "\n"}

template<typename T> struct is_plural : false_type{};
template<typename T1, typename T2> struct is_plural<pair<T1, T2>> : true_type{};
template<typename T> struct is_plural<vector<T>> : true_type{};
template<typename T> struct is_plural<complex<T>> : true_type{};
 
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << " " << p.second; }
template<typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto itr = vec.begin(); itr != vec.end(); ++itr) is >> *itr; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { if (vec.empty()) return os; bool pl = is_plural<T>(); os << vec.front(); for (auto itr = ++vec.begin(); itr != vec.end(); ++itr) os << interval[pl] << *itr; return os; }
 
bool CoutYN(bool a, string y = "Yes", string n = "No") { cout << (a ? y : n) << "\n"; return a; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

long long modpow(int a, long long n, int mod = MOD) { long long ret = 1; do { if (n & 1) ret = ret * a % mod; a = 1LL * a * a % mod; } while (n >>= 1); return ret; }

template<typename T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<typename T> T LCM(T a, T b) { return a / GCD(a, b) * b; }

template<typename T1, typename T2> bool CompareBySecond(pair<T1, T2> a, pair<T1, T2> b) { return a.second != b.second ? a.second < b.second : a.first < b.first; }


/* -------- <templates end> -------- */


void solve() {
  int n, m; cin >> n >> m;
  vi arr(n); cin >> arr;

  Graph<int> pos(m);
  rep(i,n) {
    pos[--arr[i]].emplace_back(i);
  }

  int cycle = 0;
  
  vl dec(m);
  dec[0] = 0;

  rep1(i,n) {
    if (arr[i-1] > arr[i]) {
      ++cycle;
      dec[0] += m - arr[i-1] - 1;
    }
  }

  vi cnt(m, cycle + 1);
  rep(i,arr[0]) --cnt[i];
  REP(i,arr[n-1]+1,m-1) --cnt[i];

  rep(i,m-1) {
    ll tmp = 0;

    for (auto &v : pos[i]) {
      if (v != 0) {
        int dif = i - arr[v-1];
        if (dif < 0) dif += m;
        tmp += dif - 1;
      }
    }

    dec[i+1] = dec[i] - tmp + cnt[i+1] - pos[i].size();
  }

  ll ans = 1LL * cycle * m - arr[0] + arr[n-1] - *max_element(ALL(dec));
  cout << ans << endl;
}


/* -------- <programs end> -------- */


signed main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  solve();
  return 0;
}