Contest T-shirts

Segtree has $ M $ contest T-shirts.

He decided to spend $ N $ days on the contest T-shirt alone, and told $ i = 1, 2, 3, \ dots, N $ "$ A_i $ T-shirt on the $ i $ day." I made a plan for $ N $ to wear.

However, if you keep the current plan, you may not be able to do the laundry in time, so I would like to change the plan if necessary so that I will not wear the same clothes for two consecutive days.

Find the minimum number of plans that need to be changed. It can be proved that the conditions can always be met by changing the plan under the given constraints.

input

Input is given from standard input in the following format.


$ M $ $ N $
$ A_1 $ $ A_2 $ $ \ ldots $ $ A_N $


output

Output the minimum number of plans that need to be changed.

However, insert a line break at the end.

Constraint

* $ 2 \ leq M \ leq 10 ^ 9 $
* $ 1 \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq A_i \ leq M $
* All inputs are integers.



Input example 1


twenty three
2 2 1


Output example 1


1


Input example 2


3 6
1 1 1 2 2 3


Output example 2


2






Example

Input

2 3
2 2 1


Output

1
#include <bits/stdc++.h>
#define whlie while
#define pb push_back
#define eb emplace
#define fi first
#define se second
#define rep(i,N) for(int i = 0; i < (N); i++)
#define repr(i,N) for(int i = (N) - 1; i >= 0; i--)
#define rep1(i,N) for(int i = 1; i <= (N) ; i++)
#define repr1(i,N) for(int i = (N) ; i > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; using vl = vector<ll>;
using vi = vector<int>; using vvi = vector< vector<int> >;
constexpr int inf = 1001001001;
constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(10);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);cout << endl;} while(0)
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif
using P = pair<int,int>; using vp = vector<P>;
constexpr int MOD = /** 1000000007; //*/ 998244353;
////////////////



void solve(){
  ini(M , N);
  vi a(N); in(a);
  if(M == 2){
    // 1 2 1 2
    int c1 = 0 , c2 = 0;
    rep(i , N){
      if(a[i] == 1 + (i & 1)) c1++;
      if(a[i] == 2 - (i & 1)) c2++;
    }
    trc(c1 , c2);
    out(min(c1 , c2));
    exit(0);
  }

  int ans = 0;
  for(int i = 0 ; i < N - 1; ){
    if(a[i] == a[i + 1]){
      // i + 1を変える
      ans++;
      i += 2;
    }
    else i++;
  }
  out(ans);

}
