Problem:
Score : 200 points

Problem Statement

Takahashi lives in another world. There are slimes (creatures) of 10000 colors in this world. Let us call these colors Color 1, 2, ..., 10000.

Takahashi has N slimes, and they are standing in a row from left to right. The color of the i-th slime from the left is a_i.
If two slimes of the same color are adjacent, they will start to combine themselves. Because Takahashi likes smaller slimes, he has decided to change the colors of some of the slimes with his magic.

Takahashi can change the color of one slime to any of the 10000 colors by one spell.
How many spells are required so that no slimes will start to combine themselves?

Constraints

2 \leq N \leq 100

1 \leq a_i \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of spells required.

Sample Input 1

5
1 1 2 2 2

Sample Output 1

2

For example, if we change the color of the second slime from the left to 4, and the color of the fourth slime to 5, the colors of the slimes will be 1, 4, 2, 5, 2, which satisfy the condition.

Sample Input 2

3
1 2 1

Sample Output 2

0

Although the colors of the first and third slimes are the same, they are not adjacent, so no spell is required.

Sample Input 3

5
1 1 1 1 1

Sample Output 3

2

For example, if we change the colors of the second and fourth slimes from the left to 2, the colors of the slimes will be 1, 2, 1, 2, 1, which satisfy the condition.

Sample Input 4

14
1 2 2 3 3 3 4 4 4 4 1 2 3 4

Sample Output 4

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
#define repi(i,a,b) for(int i=(a), i##_len=(b); i<i##_len; ++i)
#define rep(i,n) repi(i,0,n)
#define repi_(i,a,b) for(int i=(a), i##_len=(b); i<=i##_len; ++i)
#define rep_(i,n) repi_(i,0,n)
#define all(x) (x).begin(),(x).end()
#define vsort(v) sort((v).begin(), (v).end())
#define gsort(v) tsort(v); reverse((v).begin(), (v).end())
#define vrev(v) reverse((v).begin(), (v).end())
#define siz(v) ((int)(v).size())
#define ers(v, n) (v).erase((v).begin() + n)
#define cnt(v, n) count(all(v), n)
#define vmin(v) *min_element(v.begin(), v.end())
#define vmax(v) *max_element(v.begin(), v.end())
#define el "\n"
#define co(x) cout<<(x)<<el
#define coc(c, a, b) if(c)co((a));else co((b))
#define cosp(x) cout<<(x)<<' '
#define YES(c) coc(c,"YES", "NO")
#define Yes(c) coc(c,"Yes", "No")
#define yes(c) coc(c,"yes", "no")
#define inf INT_MAX
#ifdef tqktmp_2
#define gc() getchar(); getchar()
#else
#define gc() 1
#endif
typedef long long lint;

lint dx[8] = { 0,1,1,1,0,-1,-1,-1 },
dy[8] = { 1,1,0,-1,-1,-1,0,1 };

template<class T>inline bool maxi(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>inline bool mini(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(0);

	lint n, in; cin >> n;
	vector<int> a;
	rep(i, n) {
		cin >> in;
		a.push_back(in);
	}
	lint p = 0, ctr = 0;
	while (p < n - 2) {
		if (a[p] == a[p + 1]) {
			if (a[p] == 1) {
				if (a[p + 2] == 2)a[p + 1] = 3;
				else a[p + 1] = 2;
			}
			else if (a[p] == 2) {
				if (a[p + 2] == 1)a[p + 1] = 3;
				else a[p + 1] = 1;
			}
			else a[p + 1] = 1;
			p++; ctr++;
		}
		p++;
	}
	if (a[siz(a) - 1] == a[siz(a) - 2]) {
		ctr++;
	}
	co(ctr);
	gc();
}