description: You are given a keyboard that consists of 26 keys. The keys are arranged sequentially in one row in a certain order. Each key corresponds to a unique lowercase Latin letter.
  
  You have to type the word s on this keyboard. It also consists only of lowercase Latin letters.
  
  To type a word, you need to type all its letters consecutively one by one. To type each letter you must position your hand exactly over the corresponding key and press it.
  
  Moving the hand between the keys takes time which is equal to the absolute value of the difference between positions of these keys (the keys are numbered from left to right). No time is spent on pressing the keys and on placing your hand over the first letter of the word.
  
  For example, consider a keyboard where the letters from 'a' to 'z' are arranged in consecutive alphabetical order. The letters 'h', 'e', 'l' and 'o' then are on the positions 8, 5, 12 and 15, respectively. Therefore, it will take |5 - 8| + |12 - 5| + |12 - 12| + |15 - 12| = 13 units of time to type the word "hello". 
  
  Determine how long it will take to print the word s.
  
  Input
  
  The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.
  
  The next 2t lines contain descriptions of the test cases.
  
  The first line of a description contains a keyboard — a string of length 26, which consists only of lowercase Latin letters. Each of the letters from 'a' to 'z' appears exactly once on the keyboard.
  
  The second line of the description contains the word s. The word has a length from 1 to 50 letters inclusive and consists of lowercase Latin letters.
  
  Output
  
  Print t lines, each line containing the answer to the corresponding test case. The answer to the test case is the minimal time it takes to type the word s on the given keyboard.
  
  Example
  
  Input
  
  
  5
  abcdefghijklmnopqrstuvwxyz
  hello
  abcdefghijklmnopqrstuvwxyz
  i
  abcdefghijklmnopqrstuvwxyz
  codeforces
  qwertyuiopasdfghjklzxcvbnm
  qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
  qwertyuiopasdfghjklzxcvbnm
  abacaba
  
  
  Output
  
  
  13
  0
  68
  0
  74
solution: #include <bits/stdc++.h>
  using namespace std;
  #pragma GCC target( \
      "sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  const long long inf = 1e18;
  const double pi = 3.14159265358979323846264338327950288419716939937510;
  const long long mod = 1e9 + 7;
  const long long mod2 = 998244353;
  const double eps = (0.00000000001);
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    a %= b;
    return gcd(b, a);
  }
  long long gcdExtended(long long, long long, long long*, long long*);
  long long gcdExtended(long long a, long long b, long long* x, long long* y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    long long x1, y1;
    long long gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  long long powerForInverse(long long x, long long y, long long m) {
    if (y == 0) return 1;
    long long temp = powerForInverse(x, y / 2, m);
    temp = (temp * temp) % m;
    if (y & 1) temp = ((temp % m) * (x % m)) % m;
    return temp;
  }
  long long modInverse(long long a, long long m) {
    return powerForInverse(a, m - 2, m);
  }
  long long nCr(long long n, long long r) {
    if (r > n) return 0;
    if (r > n - r) r = n - r;
    long long ans = 1;
    for (long long i = 1; i < r + 1; i++) {
      ans *= (n - i + 1);
      ans *= modInverse(i, mod);
    }
    return ans;
  }
  long long nCrM(long long n, long long r) {
    if (r > n) return 0;
    if (r > n - r) r = n - r;
    long long ans = 1;
    for (long long i = 1; i < r + 1; i++) {
      ans *= (n - i + 1);
      ans %= mod;
      ans *= modInverse(i, mod);
      ans %= mod;
    }
    return ans;
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  long long binpowM(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a % mod;
    long long res = binpowM(a, b / 2) % mod;
    if (b % 2 == 1)
      return (((res * res) % mod) * a) % mod;
    else
      return (res * res) % mod;
  }
  long long binpow(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long res = binpow(a, b / 2);
    if (b % 2 == 1)
      return (((res * res)) * a);
    else
      return (res * res);
  }
  void sieve(vector<long long>& primes, long long n, vector<long long>& touched) {
    primes.clear();
    long long arr[n + 1];
    for (long long i = 0; i < n + 1; i++) arr[i] = i;
    touched[0] = true;
    touched[1] = true;
    for (long long i = 2; i < (int)sqrt(n) + 1; i++) {
      if (!touched[i]) {
        touched[i] = true;
        primes.push_back(i);
        for (long long j = i * i; j < n + 1; j++) {
          touched[j] = true;
          j--;
          j += i;
        }
      }
    }
    for (long long i = (int)sqrt(n) + 1; i < n; i++) {
      if (!touched[i]) primes.push_back(i);
    }
  }
  long long binary_search(long long arr[], long long n, long long elem) {
    long long l = 0, r = n - 1;
    while (r >= l) {
      long long mid = (l + r) / 2;
      if (arr[mid] > elem)
        r = mid - 1;
      else if (arr[mid] < elem)
        l = mid + 1;
      else
        return mid;
    }
    return -1;
  }
  long long binary_search_vector(vector<long long>& arr, long long elem) {
    long long n = arr.size();
    long long l = 0, r = n - 1;
    while (r >= l) {
      long long mid = (l + r) / 2;
      if (arr[mid] > elem)
        r = mid - 1;
      else if (arr[mid] < elem)
        l = mid + 1;
      else
        return mid;
    }
    return -1;
  }
  long long factorial(long long n) {
    long long res = 1;
    for (long long i = 2; i < n + 1; i++) res *= i;
    return res;
  }
  long long factorialM(long long n) {
    long long res = 1;
    for (long long i = 2; i < n + 1; i++) res = (res * i) % mod;
    return res % mod;
  }
  bool checkPrime(long long n) {
    long long count = 0;
    for (long long i = 2; i < 1 + (long long)sqrt(n); i++) {
      if (n % i == 0) count++;
    }
    if (count > 0)
      return false;
    else
      return true;
  }
  bool compare(pair<long long, long long> p1, pair<long long, long long> p2) {
    if (p1.second > p2.second)
      return true;
    else if (p1.second == p2.second)
      return p1.first < p2.first;
    else
      return false;
  }
  class Graph {
   public:
    vector<vector<long long> > adj;
    vector<bool> visited;
    Graph(long long n) {
      adj.resize(n + 1);
      visited.resize(n + 1);
    }
    void addEdge(long long first, long long second, bool directed = false);
    void dfs(long long start);
    void check(long long ans[], long long n);
  };
  void Graph::addEdge(long long first, long long second, bool directed) {
    adj[first].push_back(second);
    if (!directed) adj[second].push_back(first);
  }
  void Graph::dfs(long long start) {
    visited[start] = true;
    cout << start << " ";
    for (long long i = 0; i < adj[start].size(); i++) {
      if (!visited[adj[start][i]]) dfs(adj[start][i]);
    }
  }
  void Graph::check(long long ans[], long long n) {
    for (long long i = 0; i < adj.size(); i++) {
      ans[adj[i].size()]++;
    }
  }
  long long noOfDigits(long long x) {
    long long digits = 0;
    while (x > 0) {
      x /= 10;
      digits++;
    }
    return digits;
  }
  long long sumOfDigits(long long n) {
    long long sum = 0;
    while (n > 0) {
      sum += n % 10;
      n /= 10;
    }
    return sum;
  }
  bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
    return a.first < b.first;
  }
  void reverse(string& str, long long start, long long end) {
    for (long long i = start; i < start + ((end - start + 1) / 2); i++) {
      swap(str[i], str[end + start - i]);
    }
  }
  long long findGCD(long long arr[], long long n) {
    long long result = arr[0];
    for (int i = 1; i < n; i++) {
      result = gcd(arr[i], result);
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  long long findGCD1(long long arr[], long long n, long long i1, long long i2) {
    long long result;
    if (i1 == 0) {
      result = arr[0] * arr[i2];
    } else {
      result = arr[0];
    }
    for (int i = 1; i < n; i++) {
      if (i == i1) {
        result = gcd(arr[i] * arr[i2], result);
      } else if (i == i2) {
        result = gcd(arr[i] * arr[i1], result);
      } else {
        result = gcd(arr[i], result);
      }
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    srand(time(NULL));
    long long t;
    cin >> t;
    for (long long tc = 1; tc <= t; tc++) {
      string order;
      cin >> order;
      string word;
      cin >> word;
      long long arr[26];
      for (long long i = 0; i < 26; i++) {
        arr[(int)(order[i] - 97)] = i;
      }
      long long ans = 0;
      for (long long i = 0; i < word.length() - 1; i++) {
        long long c1 = (int)(word[i] - 97);
        long long c2 = (int)(word[i + 1] - 97);
        ans += abs(arr[c2] - arr[c1]);
      }
      cout << ans << "\n";
    }
    return 0;
  }
  