Codeforces Round 344 (Div. 2) |
---|

Finished |

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

data structures

hashing

implementation

string suffix structures

strings

*2100

No tag edit access

The problem statement has recently been changed. View the changes.

×
D. Messenger

time limit per test

2 secondsmemory limit per test

512 megabytesinput

standard inputoutput

standard outputEach 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 (*l*_{i}, *c*_{i}), where *l*_{i} is the length of the *i*-th block and *c*_{i} 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 *t*_{p}*t*_{p + 1}...*t*_{p + |s| - 1} = *s*, where *t*_{i} 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 "*l*_{i}-*c*_{i}" (1 ≤ *l*_{i} ≤ 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 "*l*_{i}-*c*_{i}" (1 ≤ *l*_{i} ≤ 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.

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: May/19/2024 10:13:26 (h1).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|