总结
第一步
把相同字母的下标放在一个一维数组
第二步
二分跑是否存在即可
/*
____________ ______________ __
/ _________ /\ /_____ _____/\ / /\
/ /\ / / \\ / /\ \ \ / / \
/ / \_____/ / / \__/ / \____\/ / / /
/ / / / / / / / / / / /
/ / / / / / / / / / / /
/ / / / / / / / / / / /
/ /___/____/ / / / / / / /___/________
/____________/ / /__/ / /______________/\
\ \ / \ \ / \ \ \
\____________\/ \__\/ \______________\/
___ ___ ___ __________
/ /\ / /\ / /\ /_______ /\
/ /__\___/ / \ / / \ \ / / \
/____ ____/ / / / / \____/ / /
\ / /\ \ / / / / / / /
\_/ / \___\/ ___ / / / / / /
/ / / / /\ / / / / / /
/ / / / /__\__/ / / / /___/____
/___/ / /___________/ / /___________/\
\ \ / \ \ / \ \ \
\___\/ \___________\/ \___________\/
A CODE OF CBOY
*/
#include<bits/stdc++.h>
//typedef long long ll;
//#define ull unsigned long long
//#define int long long
#define F first
#define S second
#define endl "\n"//<<flush
#define lowbit(x) (x&(-x))
#define ferma(a,b) pow(a,b-2)
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()
#define memset(a,b) memset(a,b,sizeof(a));
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int MAXN=0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
void file()
{
#ifdef ONLINE_JUDGE
#else
freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
#endif
}
int vis[26];
signed main()
{
IOS;
//file();
int t;
cin>>t;
while(t--)
{
vector<int>vec[26],vis(26);
string s,t;
cin>>s>>t;
for(auto &it:s)
vec[it-'a'].pb(&it-&*s.begin());
for(int i=0;i<26;i++)
vis[i]=vec[i].size();
int flag=0,ans=0,index=-1;
for(int i=0;i<t.size();i++)
{
int num=t[i]-'a';
int pos=upper_bound(all(vec[num]),index)-vec[num].begin();
if(pos==vis[num]&&index==-1)
{
flag=1;
break;
}
else if(pos==vis[num])
{
index=-1;
ans++;
i--;
}
else
{
index=vec[num][pos];
}
}
if(flag)
cout<<-1<<endl;
else
cout<<ans+1<<endl;
}
return 0;
}