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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define _p(...) (void)printf(__VA_ARGS__)
#define forr(x,arr) for(auto&& x:arr)
#define _overload3(_1,_2,_3,name,...) name
#define _rep2(i,n) _rep3(i,0,n)
#define _rep3(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,_rep3,_rep2,)(__VA_ARGS__)
#define _rrep2(i,n) _rrep3(i,0,n)
#define _rrep3(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define bit(n) (1LL<<(n))
#define sz(x) ((int)(x).size())
#define TEN(n) ((ll)(1e##n))
#define fst first
#define snd second

string DBG_DLM(int &i){return(i++==0?"":", ");}
#define DBG_B(exp){int i=0;os<<"{";{exp;}os<<"}";return os;}
template<class T>ostream&operator<<(ostream&os,vector<T>v);
template<class T>ostream&operator<<(ostream&os,set<T>v);
template<class T>ostream&operator<<(ostream&os,queue<T>q);
template<class T>ostream&operator<<(ostream&os,priority_queue<T>q);
template<class T,class K>ostream&operator<<(ostream&os,pair<T,K>p);
template<class T,class K>ostream&operator<<(ostream&os,map<T,K>mp);
template<class T,class K>ostream&operator<<(ostream&os,unordered_map<T,K>mp);
template<int I,class TPL>void DBG(ostream&os,TPL t){}
template<int I,class TPL,class H,class...Ts>void DBG(ostream&os,TPL t){os<<(I==0?"":", ")<<get<I>(t);DBG<I+1,TPL,Ts...>(os,t);}
template<class T,class K>void DBG(ostream&os,pair<T,K>p,string delim){os<<"("<<p.first<<delim<<p.second<<")";}
template<class...Ts>ostream&operator<<(ostream&os,tuple<Ts...>t){os<<"(";DBG<0,tuple<Ts...>,Ts...>(os,t);os<<")";return os;}
template<class T,class K>ostream&operator<<(ostream&os,pair<T,K>p){DBG(os,p,", ");return os;}
template<class T>ostream&operator<<(ostream&os,vector<T>v){DBG_B(forr(t,v){os<<DBG_DLM(i)<<t;});}
template<class T>ostream&operator<<(ostream&os,set<T>s){DBG_B(forr(t,s){os<<DBG_DLM(i)<<t;});}
template<class T>ostream&operator<<(ostream&os,queue<T>q){DBG_B(for(;q.size();q.pop()){os<<DBG_DLM(i)<<q.front();});}
template<class T>ostream&operator<<(ostream&os,priority_queue<T>q){DBG_B(for(;q.size();q.pop()){os<<DBG_DLM(i)<<q.top();});}
template<class T,class K>ostream&operator<<(ostream&os,map<T,K>m){DBG_B(forr(p,m){os<<DBG_DLM(i);DBG(os,p,"->");});}
template<class T,class K>ostream&operator<<(ostream&os,unordered_map<T,K>m){DBG_B(forr(p,m){os<<DBG_DLM(i);DBG(os,p,"->");});}
#define DBG_OVERLOAD(_1,_2,_3,_4,_5,_6,macro_name,...)macro_name
#define DBG_LINE(){char s[99];sprintf(s,"line:%3d | ",__LINE__);cerr<<s;}
#define DBG_OUTPUT(v){cerr<<(#v)<<"="<<(v);}
#define DBG1(v,...){DBG_OUTPUT(v);}
#define DBG2(v,...){DBG_OUTPUT(v);cerr<<", ";DBG1(__VA_ARGS__);}
#define DBG3(v,...){DBG_OUTPUT(v);cerr<<", ";DBG2(__VA_ARGS__);}
#define DBG4(v,...){DBG_OUTPUT(v);cerr<<", ";DBG3(__VA_ARGS__);}
#define DBG5(v,...){DBG_OUTPUT(v);cerr<<", ";DBG4(__VA_ARGS__);}
#define DBG6(v,...){DBG_OUTPUT(v);cerr<<", ";DBG5(__VA_ARGS__);}
#define DEBUG0(){DBG_LINE();cerr<<endl;}
#ifdef LOCAL
#define out(...){DBG_LINE();DBG_OVERLOAD(__VA_ARGS__,DBG6,DBG5,DBG4,DBG3,DBG2,DBG1)(__VA_ARGS__);cerr<<endl;}
#else
#define out(...)
#endif

using ll=long long;
using pii=pair<int,int>;using pll=pair<ll,ll>;using pil=pair<int,ll>;using pli=pair<ll,int>;
using vs=vector<string>;using vvs=vector<vs>;using vvvs=vector<vvs>;
using vb=vector<bool>;using vvb=vector<vb>;using vvvb=vector<vvb>;
using vi=vector<int>;using vvi=vector<vi>;using vvvi=vector<vvi>;
using vl=vector<ll>;using vvl=vector<vl>;using vvvl=vector<vvl>;
using vd=vector<double>;using vvd=vector<vd>;using vvvd=vector<vvd>;
using vpii=vector<pii>;using vvpii=vector<vpii>;using vvvpii=vector<vvpii>;
template<class T>bool amax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>bool amin(T&a,const T&b){return a>b?a=b,1:0;}
ll ri(){ll l;cin>>l;return l;} string rs(){string s;cin>>s;return s;}

ll solve() {
  int n = ri();
  int m = ri();

  vector<int> A(n);
  rep(i, n) A[i] = ri();

  vl ans_u(m+1, 1e18);
  {
    ll sum = 0;
    vpii U;

    rep(i, 1, n) {
      int x = A[i-1];
      int y = A[i];
      if (y > x) {
        sum += y - x;
        U.emplace_back(x, y);
      }
    }
    sort(all(U));
    out(U);
    out(sum);

    priority_queue<pii, vector<pii>, greater<pii> > q;
    int idx = 0;
    ll efe = 0;

    rep(f, 1, m+1) {
      while (idx < sz(U) && U[idx].fst < f) {
        // 効果ありになる
        int x = U[idx].fst;
        int y = U[idx].snd;
        q.emplace(y, x);
        idx++;
      }
      while (!q.empty() && q.top().fst < f) {
        // 効果なくなる
        int x = q.top().snd;
        q.pop();
        efe -= f - x - 1;
      }
      efe += sz(q);
      out(f, sum, efe, sz(q));
      ans_u[f] = sum - efe + sz(q);
    }
  }
  out(ans_u);

  vl ans_d(m+1, 1e18);
  {
    ll sum = 0;
    vpii D;
    ll efe = 0;

    rep(i, 1, n) {
      int x = A[i-1];
      int y = A[i];
      if (y < x) {
        sum += m-x+y;
        D.emplace_back(y, x);
        efe += m-x;
      }
    }
    sort(all(D));
    out(D);
    out(sum);

    int idx = 0;
    priority_queue<pii, vector<pii>, greater<pii> > q;
    int cnt = sz(D);  // 効果ありのcnt

    rep(f, 1, m+1) {
      while (idx < sz(D) && D[idx].fst < f) {
        // 効果なしになる
        int y = D[idx].fst;
        int x = D[idx].snd;
        out(y, x);
        q.emplace(x, y);
        idx++;
        cnt--;
        // y まで1だったのが、普通に処理しないといけない
        efe -= (m-x+y);
      }
      while (!q.empty() && q.top().fst < f) {
        // 効果ありに戻る
        int x = q.top().fst;
        out(q.top());
        q.pop();
        cnt++;
        efe += f - x - 1;
      }
      efe += cnt;
      out(f, sum, efe, cnt);
      ans_d[f] = sum - efe + cnt;
    }
  }
  out(ans_d);

  ll ans = 1e18;
  rep(f, 1, m+1) {
    amin(ans, ans_u[f]+ans_d[f]);
  }

  return ans;
}

void Main() {
  cout << solve() << endl;
}

signed main() { cin.tie(nullptr); ios::sync_with_stdio(false); Main(); return 0; }