Codeforces Round #344 (Div. 2)D. Messenger【kmp】

D. Messenger

time limit per test
2 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

Each employee of the "Blake Techologies" company uses a special messaging app "Blake Messenger". All the stuff likes this app and uses it constantly. However, some important futures are missing. For example, many users want to be able to search through the
message history. It was already announced that the new feature will appear in the nearest update, when developers faced some troubles that only you may help them to solve.

All the messages are represented as a strings consisting of only lowercase English letters. In order to reduce the network load strings are represented in the special compressed form. Compression algorithm works as follows: string is represented as a concatenation
of nblocks, each block containing only equal characters. One block may be described as a pair (li, ci),
where li is
the length of the i-th block and ci is
the corresponding letter. Thus, the string s may be written as the sequence of pairs 

.

Your task is to write the program, that given two compressed string t and s finds
all occurrences of s in t.
Developers know that there may be many such occurrences, so they only ask you to find the number of them. Note that p is
the starting position of some occurrence of sin t if
and only if tptp + 1...tp + |s| - 1 = s,
where ti is
the i-th character of string t.

Note that the way to represent the string in compressed form may not be unique. For example string "aaaa" may be given as 

, 

, 

...

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) —
the number of blocks in the strings t and s,
respectively.

The second line contains the descriptions of n parts of string t in
the format "li-ci"
(1 ≤ li ≤ 1 000 000) —
the length of the i-th part and the corresponding lowercase English letter.

The second line contains the descriptions of m parts of string s in
the format "li-ci"
(1 ≤ li ≤ 1 000 000) —
the length of the i-th part and the corresponding lowercase English letter.

Output

Print a single integer — the number of occurrences of s in t.

Examples

input

5 3
3-a 2-b 4-c 3-a 2-c
2-a 2-b 1-c

output

1

input

6 1
3-a 6-b 7-a 4-c 8-e 2-a
3-a

output

6

input

5 5
1-h 1-e 1-l 1-l 1-o
1-w 1-o 1-r 1-l 1-d

output

0

Note

In the first sample, t = "aaabbccccaaacc",
and string s = "aabbc". The only
occurrence of string s in string t starts
at position p = 2.

In the second sample, t = "aaabbbbbbaaaaaaacccceeeeeeeeaa",
and s = "aaa". The occurrences
of s in t start
at positions p = 1,p = 10, p = 11, p = 12, p = 13 and p = 14.

题意：给出两个字符串所给的形式为 每个字符串的每段的字符和它的长度 求子串在主串中出现的次数

解题思路：去掉子串的头和尾后kmp计算中间部分在主串中出现的次数最后判断即可

/* ***********************************************
Author : ryc
Created Time : 2016-08-16 Tuesday
File Name : E:\acm\codeforces\344D.cpp
Language : c++
Copyright 2016 ryc All Rights Reserved
************************************************ */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<list>
#include<queue>
#include<vector>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=200010;
struct Node{
long long l;
char c;
}S[maxn],t[maxn],s[maxn];
int Next[maxn];
void getNext(int n){
int i=0,j=-1;Next[i]=j;
while(i<n-1){
if(j==-1||(s[i+1].c==s[j+1].c&&s[i+1].l==s[j+1].l)){
i++;j++;
Next[i]=j;
}
else j=Next[j];
}
}
LL KMP(int n,int m){
LL ans=0;int i=1,j=0;
while(i<n-1){
if(j==-1||(t[i].c==s[j+1].c&&t[i].l==s[j+1].l))++i,++j;
else {
j=Next[j];
}
if(j+1==m-1){
if(t[i-m+1].c==s[0].c&&t[i].c==s[m-1].c&&t[i-m+1].l>=s[0].l&&t[i].l>=s[m-1].l)ans++;
}
}
return ans;
}
int main()
{
int n,m,cntn=0,cntm=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;++i){
scanf("%lld-%c",&t[cntn].l,&t[cntn].c);
if(cntn&&t[cntn].c==t[cntn-1].c){
t[cntn-1].l+=t[cntn].l;
}
else {
cntn++;
}
}
for(int i=1;i<=m;++i){
scanf("%lld-%c",&s[cntm].l,&s[cntm].c);
if(cntm&&s[cntm].c==s[cntm-1].c){
s[cntm-1].l+=s[cntm].l;
}
else {
cntm++;
}
}
LL ans=0;
if(cntm==1){
for(int i=0;i<cntn;i++){
if(s[0].c==t[i].c&&s[0].l<=t[i].l)ans+=t[i].l-s[0].l+1ll;
}
}
else if(cntm==2){
for(int i=0;i<cntn;i++){
if(s[0].c==t[i].c&&s[1].c==t[i+1].c&&s[0].l<=t[i].l&&s[1].l<=t[i+1].l)
ans++;
}
}
else {
getNext(cntm);
ans=KMP(cntn,cntm);
}
printf("%lld\n",ans);
return 0;
}