#include <bits/stdc++.h>
using namespace std;
#define dbg(x) x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define pv2(x) dbg(cerr << #x << " = " << x.first << ',' << x.second << '\n')
#define parr(x) dbg(prt(#x << " = { "); for (auto y : x) prt(y << ' '); prt("}\n");)
#define parr2(x) dbg(prt(#x << " = { "); for (auto [y, z] : x) prt(y << ',' << z << " "); prt("}\n");)
#define parr2d(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr(arr);} prt('\n'));
#define parr2d2(x) dbg(prt(#x << ":\n"); for (auto arr : x) {parr2(arr);} prt('\n'));
/*
build 1 or 2 bridges
add to baseline of sum of all distances driven without crossing bridge + n
then only consider people crossing
k=1:
minimize the sum of all abs(s[i] - x) + abs(t[i] - x)
one of the given locs is optimal
so here treat all s[i] and t[i] the same, sort, etc.
k=2:
if l <= x or y <= r, dist = r - l
else, dist = r - l + 2 * (dist btwn either end & closest bridge)
assume x < y
5 cases
1) left of x
2) contains x
3) btwn x, y
4) contains y
5) right of y
best y for each x?
everything to left/containing x is handled (i.e. everything with l <= x)
n^2 sol:
for (each i)
case 1: x = l[i]
everything true
case 2: x = r[i]
*/
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int k, n;
cin >> k >> n;
vector<array<int, 2>> a;
long long bsl = 0;
for (int i = 0; i < n; i++) {
char c1, c2; int i1, i2;
cin >> c1 >> i1 >> c2 >> i2;
if (c1 == c2) {
bsl += abs(i2 - i1);
} else {
if (i2 > i1) swap(i1, i2);
a.push_back({i1, i2});
bsl++;
}
}
n = (int) a.size();
vector<int> b;
for (int i = 0; i < n; i++) {
b.push_back(a[i][0]); b.push_back(a[i][1]);
}
sort(b.begin(), b.end());
if (k == 1) {
long long sum = 0;
for (int i = 0; i < 2 * n; i++) {
sum += b[i] - b[0];
}
long long best = sum;
for (int i = 1; i < 2 * n; i++) {
sum += (long long) i * (b[i] - b[i - 1]);
sum -= (long long) (2 * n - i) * (b[i] - b[i - 1]);
best = min(best, sum);
}
cout << best + bsl << '\n';
} else {
long long best = 1e18;
for (int x = 0; x < 2 * n; x++) {
for (int y = x + 1; y < 2 * n; y++) {
long long sum = 0;
for (int i = 0; i < n; i++) {
sum += min(abs(a[i][0] - b[x]) + abs(a[i][1] - b[x]),
abs(a[i][0] - b[y]) + abs(a[i][1] - b[y]));
}
best = min(best, sum);
}
}
cout << best + bsl << '\n';
}
}
/*
any observations help
check every line
IF YOUR LINES AREN'T WRONG
CHECK IF YOUR LINES ARE IN THE RIGHT ORDER
NEVER GIVE UP
*/
