I haven't written KMP for a long time.
It's totally unexpected.

First, merge the same continuum.

Assuming that the number of s characters is 1, it is legal as long as the length is sufficient and the characters are the same.

Assuming that the number of s characters is 2, the length of both sides is enough.
But only one position is legal and cannot be moved left or right.

Assuming that the number of s characters >= 3, the first and last two lengths are sufficient and all of them are equal in the middle.
So get your own getfail and then match it with t
It is important to note that the length of the sentence is sufficient.
So we need to change the special judgment.

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ans() printf("%d",ans)
#define ansn() printf("%d\n",ans)
#define anss() printf("%d ",ans)
#define lans() printf("%lld",ans)
#define lanss() printf("%lld ",ans)
#define lansn() printf("%lld\n",ans)
#define fansn() printf("%.10f\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define rsz(i,v) for(int i=0;i<(int)v.size();++i)
#define szz(x) ((int)x.size())
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp make_pair
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
#define sqr(a) ((a)*(a))
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000000+7;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
const ll infl = 10000000000000000;
const int maxn=  200000+10;
const int maxm = 40000+10;
//Pretests passed
int in(int &ret)
{
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}

struct node
{
    ll l;
    char ch;
    bool operator ==(const node &o)
    {
        return l==o.l&&ch==o.ch;
    }
    bool operator <= (const node &o)
    {
        return l<=o.l&&ch==o.ch;
    }
};
node t[maxn],s[maxn];
int n,m;
void init()
{
    int j = 0;
    for(int i=0;i<n;++i)
    {
        if(j==0||t[j-1].ch!=t[i].ch)t[j++] = t[i];
        else t[j-1].l+=t[i].l;
    }
    n = j;
    j = 0;
    for(int i=0;i<m;++i)
    {
        if(j==0||s[j-1].ch!=s[i].ch)s[j++] = s[i];
        else s[j-1].l+=s[i].l;
    }
    m = j;
}
int nt[maxn];
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

//    int n,m;
    sdd(n,m);
    r0(i,n) scanf("%lld-%c",&t[i].l,&t[i].ch);
    r0(i,m) scanf("%lld-%c",&s[i].l,&s[i].ch);
    init();
    ll ans = 0;
    if(m==1)
    {
        r0(i,n)
        {
            if(s[0]<=t[i])ans+=t[i].l-s[0].l+1;
        }
    }
    else if(m==2)
    {
        r0(i,n-1)
        {
            if(s[0]<=t[i]&&s[1]<=t[i+1])++ans;
        }
    }
    else
    {
        nt[1] = 0;
        for(int i = 2 ; i < m - 1 ; ++i)
        {
            int j = nt[i-1];
            while(j&&! (s[j+1]==s[i]) )j = nt[j];
            if(s[j+1]==s[i])++j;
            nt[i] = j;
        }
        for(int i = 1, j = 0 ; i<n-1;++i)
        {
            while(j&&!(s[j+1]==t[i]))j = nt[j];
            if(s[j+1]==t[i])++j;
            if(j==m-2)
            {
                if(s[0]<=t[i-j]&&s[j+1]<=t[i+1])++ans;
                j = nt[j];
            }
        }
    }
    lansn();
    return 0;
}