B. Email from Polycarp

B. Email from Polycarp

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Methodius received an email from his friend Polycarp. However, Polycarp's keyboard is broken, so pressing a key on it once may cause the corresponding symbol to appear more than once (if you press a key on a regular keyboard, it prints exactly one symbol).

For example, as a result of typing the word "hello", the following words could be printed: "hello", "hhhhello", "hheeeellllooo", but the following could not be printed: "hell", "helo", "hhllllooo".

Note, that when you press a key, the corresponding symbol must appear (possibly, more than once). The keyboard is broken in a random manner, it means that pressing the same key you can get the different number of letters in the result.

For each word in the letter, Methodius has guessed what word Polycarp actually wanted to write, but he is not sure about it, so he asks you to help him.

You are given a list of pairs of words. For each pair, determine if the second word could be printed by typing the first one on Polycarp's keyboard.

Input

The first line of the input contains one integer n descriptions of pairs.

The first line of each description contains a single non-empty word 106.

It is guaranteed that the total length of all words 106.

Output

Output NO.

Examples

input

Copy

4
hello
hello
hello
helloo
hello
hlllloo
hello
helo

output

Copy

YES
YES
NO
NO

input

Copy

5
aa
bb
codeforces
codeforce
polycarp
poolycarpp
aaaa
aaaab
abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz

output

Copy

NO
NO
YES
NO
NO

 

 

 

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
string s1, s2;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int flag = 0;
        cin >> s1 >> s2;
        int i = 0, j = 0;
        if (s1.size() > s2.size())
            flag = 1;
        else
        {
            for (; s2[j]; j++, i++)
            {
                if (s1[i] != s2[j])
                {
                    if (i == 0)
                    {
                        flag = 1;
                        break;
                    }
                    else
                    {
                        if (s2[j] == s2[j - 1])
                            i--;//如果s2有重复的字母，那么s1[i]保持不变，继续比较s[j+1]
                        else
                        {
                            flag = 1;
                            break;
                        }

                    }
                }

            }
        }
        if (flag == 0 && i == s1.size())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}