description: It is a simplified version of problem F2. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  4
  1 1
  221 2
  177890 2
  998244353 1
  
  
  Output
  
  
  1
  221
  181111
  999999999
solution: #include <bits/stdc++.h>
  bool isp(int x) {
    if (x < 2) return 0;
    for (int i = 2; i <= sqrt(x); i++) {
      if (x % i == 0) return 0;
    }
    return 1;
  }
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
  int FPW(int x, int y, int mod) {
    int res = 1;
    for (; y; y >>= 1) {
      if (y % 2) res = res * x % mod;
      x = x * x % mod;
    }
    return res;
  }
  using namespace std;
  int cnts(int x) {
    int ret = 0;
    while (x) {
      ret |= (1 << (x % 10));
      x /= 10;
    }
    return __builtin_popcount(ret);
  }
  int main() {
    int T = read();
    while (T--) {
      int N = read(), K = read();
      int ans = N;
      while (cnts(ans) > K) {
        int l = 1, r = ans;
        while (cnts(r) > K) {
          l *= 10, r /= 10;
        }
        l /= 10;
        ans /= l;
        ans++;
        ans *= l;
      }
      cout << ans << endl;
    }
    return 0;
  }
  