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>
  using namespace std;
  const long long INF = 0x7fffffffffffffff;
  const long long MAXN = 2e5 + 9;
  void fast_stream() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long len(long long n) {
    if (n == 0) return 1;
    long long ans = 0;
    while (n > 0) ans++, n /= 10;
    return ans;
  }
  long long lft(long long p, long long l1, long long l2) {
    long long l = l1 - l2;
    while (l--) p *= 10;
    return p;
  }
  long long rgh(long long k0, long long l) {
    long long res = 0;
    while (l--) res = 10 * res + k0;
    return res;
  }
  long long Ans, t;
  void dfs(long long n, long long p, long long k1, long long k2, long long l1,
           long long l2) {
    if (lft(p, l1, l2) > Ans || lft(p, l1, l2) + rgh(max(k1, k2), l1 - l2) < p)
      return;
    if (l1 == l2) {
      if (p >= n) Ans = min(Ans, p);
      return;
    }
    if (!(l2 == 0 && k1 == 0)) dfs(n, 10 * p + k1, k1, k2, l1, l2 + 1);
    if (!(l2 == 0 && k2 == 0)) dfs(n, 10 * p + k2, k1, k2, l1, l2 + 1);
  }
  void solve() {
    long long n, k;
    scanf("%lld%lld", &n, &k);
    long long lenN = len(n);
    Ans = INF;
    for (long long k1 = 0; k1 <= 9; k1++) {
      for (long long k2 = k1; k2 <= 9; k2++) {
        if (k == 1 && k1 != k2) continue;
        if (rgh(max(k2, k1), lenN) < n) continue;
        if (rgh(min(k1, k2), lenN) > Ans) continue;
        dfs(n, 0, k1, k2, lenN, 0);
      }
    }
    printf("%d\n", Ans);
  }
  signed main() {
    fast_stream();
    scanf("%lld", &t);
    while (t--) {
      solve();
    }
  }
  