Codeforces D. Messenger

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

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

Copy
1
input

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

Copy
6
input

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

Copy
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.

题目大意：每行给出的是一个字符串的每个字符的个数，按顺序给出的，让你求第二个字符串在第一个字符串中作为字符的位置有几个。

解题思路：刚开始我想先把字符串组合出来，然后用KMP去在第一个字符串上进行KMP，结果看了下数据量。。。MMP。

可以对于每个串给出的字符和对应的数去进行处理，放到结构体里，最后根据字符和个数进行模拟就可以了。

/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
//#include <iostream>
//#include <cstdio>
//#include <cstring>
//#include <cmath>
//#include <algorithm>
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 2e5 + 10 ;
const int INF = 0x3f3f3f3f ;
const double PI = acos(-1.0) ;
const ull seed = 133 ;

struct Part {
    char c;
    ll l;
    bool operator==(const Part &b) const {
        return (c == b.c) && (l == b.l);
    }
    bool operator<=(const Part &b) const {
        return (c == b.c) && (l <= b.l);
    }
};

int n, m, pi[Maxn];
Part a[Maxn], b[Maxn];
ll ans;

void compress(Part *a, int &n) {
    int m = 0;
    for (int i = 0; i < n; i++) {
        if (m == 0 || a[m - 1].c != a[i].c)
            a[m++] = a[i];
        else
            a[m - 1].l += a[i].l;
    }
    n = m;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        scanf("%I64d-%c", &a[i].l, &a[i].c);
    for (int i = 0; i < m; i++)
        scanf("%I64d-%c", &b[i].l, &b[i].c);
    compress(a, n), compress(b, m), ans = 0;
//    cout << m << endl ;
    if (m == 1) {
        for (int i = 0; i < n; i++){
            if (b[0] <= a[i]) ans += a[i].l - b[0].l + 1;
        }
    }
    else if (m == 2) {
//        cout << "++++" << endl ;
        for (int i = 0; i < n - 1; i++){
            if (b[0] <= a[i] && b[1] <= a[i + 1]) ans++;
        }
    }
    else {
        pi[1] = 0;
        for (int i = 2; i < m - 1; i++) {
            int j = pi[i - 1];
            while (j > 0 && !(b[j + 1] == b[i]))  j = pi[j];
            if (b[j + 1] == b[i]) j++;
            pi[i] = j;
        }
        for (int i = 1, j = 0; i < n - 1; i++) {
            while (j > 0 && !(b[j + 1] == a[i])) j = pi[j];
            if (b[j + 1] == a[i]) j++;
            if (j == m - 2) {
                if (b[0] <= a[i - j] && b[j + 1] <= a[i + 1]) ans++;
                j = pi[j];
            }
        }
    }
    printf("%I64d", ans);
    return 0;
}

Codeforces D. Messenger

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