CF 631D KMP运用

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 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.

分析：对模式串和原串进行预处理，因为不可能完全展开进行匹配。连续相同的需要合并，比如3-a 5-a 这样子连续的读入需要合并成一个片段。

如果模式串字符个数为1，那么只要长度足够且字符相同则合法。

如果模式串字符个数为2，那么左右两边长度足够且对应字符相同则合法。

如果模式串字符数大于2，那么需要首尾两个长度足够，并且中间字符串全部想等才合法。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int maxn = 2e5+10;
char s[maxn],t[maxn],st[maxn];
ll cnts[maxn],cntt[maxn];
ll Next[maxn];
int n,m;
ll ans;
ll lent,lens;
ll f,e;

void init(){
    ans = 0;
    lent = lens = 0;
    s[0] = t[0] = 0;

    char c[2];
    ll x;
    for (int i=1; i<=n; i++) {
         scanf("%I64d-%s",&x,c);
         if (c[0]==s[lens]) cnts[lens] +=x;
         else {
             cnts[++lens] = x;
             s[lens] = c[0];
         }
    }

    for (int i=1; i<=m; i++) {
         scanf("%I64d-%s",&x,c);
         if (c[0]==t[lent]) cntt[lent] +=x;
         else {
             cntt[++lent] = x;
             t[lent] = c[0];
         }
    }
}

void getnxt(){
    Next[0] = Next[1] = 0;
    ll j = 0;
    for (ll i=2; i<=lent-2; i++) {
        while (j && (st[j+1]!=st[i] || cntt[i]!=cntt[j+1])) j = Next[j];
        if (st[j+1] == st[i] && cntt[i] == cntt[j+1]) j++;
        Next[i] = j;
    }
}

ll kmp(){
   ll sum = 0;
   ll j = 0;
   for (ll i=2; i<lens; i++) {
      while (j && (st[j+1]!=s[i] || cnts[i]!=cntt[j+1])) j = Next[j];
      if (st[j+1] == s[i] && cnts[i] == cntt[j+1]) j++;
      if (j==lent-2){
         if (s[i-(lent-2)]==t[1] && s[i+1]==t[lent] && cnts[i-(lent-2)]>=f && cnts[i+1]>=e) sum++;
         j = Next[j];
      }
   }
   return sum;
}

int main(){
    while (scanf("%d %d",&n,&m)!=EOF){
        init();
        if (lent==1) {
            for (int i=1; i<=lens; i++){
                if (s[i]==t[lent] && cnts[i]>=cntt[lent])
                    ans += cnts[i] - cntt[lent]+1;
            }
        }
        else if (lent==2) {
            for (int i=1; i<lens; i++) {
                if (s[i]==t[1] && s[i+1]==t[2] && cnts[i]>=cntt[1] && cnts[i+1]>=cntt[2]) ans++;
            }
        }
        else {
           strcpy(st+1,t+2);
           st[lent] = 0;
           f = cntt[1] , e = cntt[lent];
           for (int i=1; i<lent; i++) cntt[i] = cntt[i+1];
           getnxt();
           ans = kmp();
        }
        printf("%I64d\n",ans);
    }
    return 0;
}