#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
// const int N=
int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int k,n;
cin>>k>>n;
vector<pair<int,int>> a;
vector<int> tp;
ll ans=0;
ll sm1=0,sm2=0,c1=0,c2=0;
ll sup=0;
for(int i=0;i<n;i++)
{
char c,cp;
int x,y;
cin>>c>>x>>cp>>y;
if(x>y)
swap(x,y);
ans+=(y-x);
if(c!=cp)
{
ans+=1;
a.push_back({x,y});
for(int p=-100;p<=100;p++)
{
tp.push_back(y+p);
tp.push_back(x+p);
tp.push_back(((x+y)/2)+p);
}
sup+=y;
sm1+=x;
c1++;
}
}
tp.push_back(sm1/c1);
tp.push_back(sm1/c1 +1);
tp.push_back(sm1/c1 -1);
tp.push_back(sup/c1);
tp.push_back(sup/c1 +1);
tp.push_back(sup/c1 -1);
// computing b[i][j] is simple if we can do k=1 in O(n)
sort(begin(a),end(a));
sort(begin(tp),end(tp));
int j=0;
set<pair<int,int>> ind;
ll mi=1e18;
for(int i=0;i<tp.size();i++)
{
while(j<a.size() and a[j].first<=tp[i])
{
sm1-=a[j].first;
c1--;
ind.insert({a[j].second,j});
j++;
}
while(ind.size()>0 and begin(ind)->first<tp[i])
{
sm2+=begin(ind)->first;
c2++;
ind.erase(begin(ind));
}
mi=min(mi,2ll*c2*tp[i]-2ll*sm2 + 2ll*sm1 -2ll*tp[i]*c1);
// We want bridge at tp[i]
// a[i].first<=tp[i]<=a[i].second zero extra cost
// tp[i]<a[i].first<=a[i].second 2*(a[i].first-tp[i])
// a[i].first<=a[i].second<tp[i] 2*(tp[i]-a[i].second)
}
cout<<ans+mi<<endl;
// a[i].first<=a[i+1].first and so on
// a[i].second is not sorted
}
