func_code_string: #include <bits/stdc++.h>   using namespace std;   bool primecheck(long long int a) {     bool check = true;     for (long long int i = 2; i <= sqrt(a); i++) {       if (a % i == 0) {         check = false;         break;       }     }     return (check);   }   vector<long long int> printDivisors(long long int n) {     vector<long long int> sol;     for (long long int i = 1; i <= sqrt(n); i++) {       if (n % i == 0) {         if (n / i == i) {           sol.push_back(i);         } else {           sol.push_back(i);           sol.push_back(n / i);         }       }     }     return (sol);   }   long long int digits(long long int a) { return (floor(log10(a)) + 1); }   long long binpow(long long a, long long b, long long m) {     a %= m;     long long res = 1;     while (b > 0) {       if (b & 1) res = res * a % m;       a = a * a % m;       b >>= 1;     }     return res;   }   int gcd(int a, int b) {     if (a == 0) return b;     if (b == 0) return a;     if (a == b) return a;     if (a > b) return gcd(a - b, b);     return gcd(a, b - a);   }   long long int lcm(long long int a, long long int b) {     return (a * b) / gcd(a, b);   }   bool check(string a, string b) {     long long int freq1[26] = {0};     long long int i;     for (i = 0; i < a.size(); i++) {       freq1[a[i] -  a ]++;     }     for (i = 0; i < b.size(); i++) {       freq1[b[i] -  a ]--;     }     bool sol = true;     for (i = 0; i < 26; i++)       if (freq1[i] != 0) sol = false;     return (sol);   }   int main() {     ios_base::sync_with_stdio(false);     cin.tie(NULL);     long long int a, b, c, i, j, k, l, m, n, t;     cin >> t;     while (t--) {       string num1, num2;       cin >> num1 >> num2;       bool sol;       long long int size1 = num1.size();       long long int size2 = num2.size();       if (size2 < size1) {         cout <<  NO               <<   n ;       } else {         sol = false;         for (i = 0; i <= size2 - size1; i++) {           string fresh = num1;           for (j = 0; j < size1; j++) {             fresh[j] = num2[i + j];           }           if (check(num1, fresh)) {             cout <<  YES                   <<   n ;             sol = true;             break;           }         }         if (!sol)           cout <<  NO                 <<   n ;       }     }   }   