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 n blocks, 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 s in 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.

【题解】

以num个X字母的形式给出某个字符串t和s;
求s在t中的匹配个数；
s最后的长度最大为20W*100W….
不能直接用KMP。
只能在“压缩”的状态下搞；
这道题提供了这类问题的解法->依然是KMP；
只是在做KMP的时候要增加判断一下这个连续的区域块是不是全都是相同的；
然后做KMP的时候，匹配要从匹配串的第二位开始匹配；然后到倒数第二个；
这一段匹配之后再比较第一位和最后一位;
因为
aaabbbbccc
和abbbbc也是匹配的；
(压缩下的KMP);
注意开Long LongTAT

#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define LL long longusing namespace std;const int MAXN = 3e5;int n,m,lent = 0,lens = 0;
LL tl[MAXN],sl[MAXN],fail[MAXN];
char tc[MAXN],sc[MAXN];const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
void input_LL(LL &r)
{r = 0;char t = getchar();while (!isdigit(t)) t = getchar();LL sign = 1;if (t == '-')sign = -1;while (!isdigit(t)) t = getchar();while (isdigit(t)) r = r * 10 + t - '0', t = getchar();r = r*sign;
}void input_int(int &r)
{r = 0;char t = getchar();while (!isdigit(t)) t = getchar();int sign = 1;if (t == '-')sign = -1;while (!isdigit(t)) t = getchar();while (isdigit(t)) r = r * 10 + t - '0', t = getchar();r = r*sign;
}int main()
{//freopen("F:\\rush.txt", "r", stdin);input_int(n);input_int(m);//...while (n--){int num;char key;scanf("%d-%c",&num,&key);if (lent>0 && tc[lent] == key)tl[lent]+=num;elsetc[++lent] = key,tl[lent]=num;}while (m--){int num;char key;scanf("%d-%c",&num,&key);if (lens>0 && sc[lens] == key)sl[lens]+=num;elsesc[++lens] = key,sl[lens]=num;}LL ans = 0;if (lens == 1){for (int i = 1;i <= lent;i++)if (tc[i] == sc[1])ans += max(0LL,tl[i]-sl[1]+1LL);}else{fail[2] = 2;fail[3] = 2;for (int i = 3;i <= lens;i++){int j = fail[i];while (j > 2 && (sc[j]!=sc[i] || sl[j]!=sl[i])) j = fail[j];fail[i+1] = (sc[j]==sc[i] && sl[j]==sl[i])?j+1:2;}int j = 2;for (int i = 1;i <=lent-1;i++){while (j > 2 && (sc[j]!=tc[i] || sl[j]!=tl[i])) j = fail[j];if (j <=lens-1 && (sc[j]==tc[i]&& sl[j]==tl[i])) j++;if (j==lens){if (sc[j] == tc[i+1] && sl[j] <= tl[i+1] && sc[1] == tc[i-lens+2] && sl[1] <= tl[i-lens+2])ans++;j = fail[j];}}}printf("%I64d\n",ans);return 0;
}