#include<bits/stdc++.h>

#define de(x) cout<<#x<<"="<<x<<endl;

#define dd(x) cout<<#x<<"="<<x<<" ";

#define rep(i,a,b) for(int i=a;i<(b);++i)

#define repd(i,a,b) for(int i=a;i>=(b);--i)

#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)

#define ll long long

#define mt(a,b) memset(a,b,sizeof(a))

#define fi first

#define se second

#define inf 0x3f3f3f3f

#define INF 0x3f3f3f3f3f3f3f3f

#define pii pair<int,int>

#define pdd pair<double,double>

#define pdi pair<double,int>

#define mp(u,v) make_pair(u,v)

#define sz(a) (int)a.size()

#define ull unsigned long long

#define ll long long

#define pb push_back

#define PI acos(-1.0)

#define qc std::ios::sync_with_stdio(false)

#define db double

#define all(a) a.begin(),a.end()

const int mod = 1e9+;

const int N = 5e3+;

const double eps = 1e-;

using namespace std;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }

bool ls(const db &a, const db &b) { return a + eps < b; }

bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }

ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };

ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }

ll kpow(ll a,ll b) {ll res=; if(b<) return ; for(;b;b>>=){if(b&)res=res*a;a=a*a;}return res;}

ll read(){

    ll x=,f=;char ch=getchar();

    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m;

int x[N],y[N],y2[N],x2[N],col[N];

int ans;

void gao(int d, int xs, int ys, int xe, int ye) {

    while(d <= m && ( xs > x2[d] || xe < x[d] || ys > y2[d] || ye < y[d] ) ) d++;

    if(d == m + ) {

        ans -= (ye - ys + ) * (xe - xs + );

        return ;

    }

    // 想想被每个后面的矩形的四条边，每条切割完后的剩下什么

    int k1 = max(xs, x[d]);

    if(xs < k1) gao(d + , xs, ys, k1 - , ye);

    int k2 = min(xe, x2[d]);

    if(xe > k2) gao(d + , k2 + , ys, xe, ye);

    xs = k1, xe = k2;

    k1 = max(ys, y[d]);

    if(ys < k1) gao(d + , xs, ys, xe, k1 - );

    k2 = min(ye, y2[d]);

    if(ye > k2) gao(d + , xs, k2 + , xe, ye);

}

int main(){

    scanf("%d%d",&n,&m); ans = n * n;

    rep(i, , m+) {

        char c;

        scanf("%d%d%d%d %c",&x[i],&y[i],&x2[i],&y2[i],&c);

        if(x[i] > x2[i]) swap(x2[i], x[i]);

        if(y[i] > y2[i]) swap(y2[i], y[i]);

        col[i] = c == 'b' ?  : ;

    }

    repd(i,m,) if(col[i]) gao(i+,x[i],y[i],x2[i],y2[i]);

    printf("%d\n",ans);

    return ;

}