/*
HDU 6170 - Two strings [ DP ] | 2017 ZJUT Multi-University Training 9
题意:
定义*可以匹配任意长度,.可以匹配任意字符,问两串是否匹配
分析:
dp[i][j] 代表B[i] 到 A[j]全部匹配
然后根据三种匹配类型分类讨论,可以从i推到i+1
复杂度O(n^2)
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 2505;
int t;
char a[N], b[N];
int la, lb;
bool dp[N][N];
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%s%s", a+1, b+1);
la = strlen(a+1);
lb = strlen(b+1);
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for (int i = 1; i <= lb; i++)
{
if (b[i] == '.')
{
for (int j = 1; j <= la; j++) dp[i][j] = dp[i-1][j-1];
}
else if (b[i] == '*')
{
for (int j = 0; j <= la; j++) dp[i][j] = dp[i-2][j];
for (int j = 0; j <= la; )
{
if (dp[i-1][j])
{
int k = j;
while (a[k] == a[j])
{
dp[i][k] = 1;
k++;
}
j = k;
}
else j++;
}
}
else
{
for (int j = 1; j <= la; j++)
if (dp[i-1][j-1] && a[j] == b[i])
dp[i][j] = 1;
}
}
if (dp[lb][la]) puts("yes");
else puts("no");
}
}