Educational Codeforces Round 81 (Rated for Div. 2)C 二分

C. Obtain The String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two strings ss and tt consisting of lowercase Latin letters. Also you have a string zz which is initially empty. You want string zz to be equal to string tt. You can perform the following operation to achieve this: append any subsequence of ss at the end of string zz. A subsequence is a sequence that can be derived from the given sequence by deleting zero or more elements without changing the order of the remaining elements. For example, if z=acz=ac, s=abcdes=abcde, you may turn zz into following strings in one operation:

1. z=acacez=acace (if we choose subsequence aceace);
2. z=acbcdz=acbcd (if we choose subsequence bcdbcd);
3. z=acbcez=acbce (if we choose subsequence bcebce).

Note that after this operation string ss doesn't change.

Calculate the minimum number of such operations to turn string zz into string tt.

Input

The first line contains the integer TT (1≤T≤1001≤T≤100) — the number of test cases.

The first line of each testcase contains one string ss (1≤|s|≤1051≤|s|≤105) consisting of lowercase Latin letters.

The second line of each testcase contains one string tt (1≤|t|≤1051≤|t|≤105) consisting of lowercase Latin letters.

It is guaranteed that the total length of all strings ss and tt in the input does not exceed 2⋅1052⋅105.

Output

For each testcase, print one integer — the minimum number of operations to turn string zz into string tt. If it's impossible print −1−1.

Example

input

Copy

3
aabce
ace
abacaba
aax
ty
yyt

output

Copy

1
-1
3

很经典的二分题目，似乎做过了很多遍

//AUTHOR : cwb
#include
using namespace std;
vector ve[30];
char s[100005];
char t[100005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int flag=1;
        int ans=1;
        scanf("%s",s+1);
        scanf("%s",t+1);
        vector ve[30];
        for(int i=1;s[i];i++)
        {
            ve[s[i]-'a'].push_back(i);
        }
        int cur=0;
        for(int i=1;t[i];i++)
        {
            int tmp=t[i]-'a';
            if(ve[tmp].size()==0)
            {
                flag=0;
                break;
            }
            else
            {
                int pos=upper_bound(ve[tmp].begin(),ve[tmp].end(),cur)-ve[tmp].begin();
                if(pos==ve[tmp].size())
                {
                    ans++;
                    pos=upper_bound(ve[tmp].begin(),ve[tmp].end(),0)-ve[tmp].begin();
                    cur=ve[tmp][pos];
                }
                else
                {
                    cur=ve[tmp][pos];
                }
            }
            //cout<