#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ss second
#define ff first
typedef vector<pair<int, int>> vpi;
typedef vector<int> vi;
void solve() {
int k, n;
cin >> k >> n;
int ans = 0;
map<int, int> mp;
vpi v;
for (int i = 0; i < n; i++) {
char a, b;
int c, d;
cin >> a >> c >> b >> d;
if (a == b)
ans += abs(d - c);
else {
if (c > d) swap(c, d);
mp[c]++;
mp[d + 1]--;
v.push_back({c, d});
}
}
vi t;
int mx = 0, p = 0;
while (k--) {
int cur = 0;
for (auto u : mp) {
cur += u.ss;
if (mx <= cur) {
mx = cur;
p = u.ff;
}
}
for (auto u : v) {
if (u.ff <= p && u.ss >= p) {
mp[u.ff]--;
mp[u.ss + 1]++;
}
}
t.push_back(p);
mx = 0;
p = 0;
}
for (auto u : v) {
int mn = 1e10;
for (auto j : t) {
mn = min(mn, abs(j - u.ff) + 1 + abs(j - u.ss));
}
ans += mn;
}
cout << ans << "\n";
}
/*
Fact 1: add a bridge between office and home and add a bridge at the end of office or home are the same -> just add at the end
Fact 2: if pair a is inside pair b, ignore pair b
Fact 3: ignore pair on the same side
Fact 4: it is optimal to choose a point with maximum pair included
*/
/*
Step:
1. Find the point with maximum number of pair included in each loop
2. erase all pair included in the point
3. Calculate the minimum cost after k point is used
time: O(n*log(n))
*/
signed main() {
int t = 1;
// cin>>t;
while (t--) {
solve();
}
}
