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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
#include<numeric>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const int mod=1000000007;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-4;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(const vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(const vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
int main(void){
	int i,m,n,ans=0;cin>>m>>n;
	
	if(m==2){
		int ki=0,gu=0;
		for(i=0;i<n;i++){
			int a;cin>>a;
			if((a+i)%2==0){ki++;}
			else{gu++;}
		}
		cout<<min(ki,gu)<<endl;
	}else{
		int mae=0;
		for(i=0;i<n;i++){
			int a;cin>>a;
			if(mae==a){a=0;ans++;}
			mae=a;
		}
		cout<<ans<<endl;
	}
	
	return 0;
}
