#include <bits/stdc++.h>
using namespace std;
int t, n, k;
int a[20], al;
int b[20], bl;
int d[20];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &k);
    memset(d, 0, sizeof(d));
    memset(a, 0, sizeof(a));
    memset(b, -1, sizeof(b));
    al = 0;
    while (n) {
      a[al] = n % 10;
      n /= 10;
      al++;
    }
    a[al] = 0;
    bl = al;
    int l = bl - 1, c = 0;
    while (l != -1) {
      if (d[a[l]] == 0) {
        c++;
        if (c > k) {
          c = k;
          break;
        }
      }
      d[a[l]]++;
      b[l] = a[l];
      l--;
    }
    if (l != -1) {
      bool tf = false;
      while (1) {
        for (int i = a[l] + 1; i <= 9; i++) {
          if ((d[i] != 0) || (c < k)) {
            if (d[i] == 0) c++;
            b[l] = i;
            d[i]++;
            tf = true;
            break;
          }
        }
        if (tf) break;
        l++;
        d[a[l]]--;
        if (d[a[l]] == 0) c--;
      }
      int m = 0;
      if (c == k) {
        for (int i = 0; i <= 9; i++)
          if (d[i] != 0) {
            m = i;
            break;
          }
      }
      for (int i = l - 1; i >= 0; i--) b[i] = m;
    }
    for (int i = bl - 1; i >= 0; i--) printf("%d", b[i]);
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int t, n, k;
int a[20], al;
int b[20], bl;
int d[20];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d %d", &n, &k);
    memset(d, 0, sizeof(d));
    memset(a, 0, sizeof(a));
    memset(b, -1, sizeof(b));
    al = 0;
    while (n) {
      a[al] = n % 10;
      n /= 10;
      al++;
    }
    a[al] = 0;
    bl = al;
    int l = bl - 1, c = 0;
    while (l != -1) {
      if (d[a[l]] == 0) {
        c++;
        if (c > k) {
          c = k;
          break;
        }
      }
      d[a[l]]++;
      b[l] = a[l];
      l--;
    }
    if (l != -1) {
      bool tf = false;
      while (1) {
        for (int i = a[l] + 1; i <= 9; i++) {
          if ((d[i] != 0) || (c < k)) {
            if (d[i] == 0) c++;
            b[l] = i;
            d[i]++;
            tf = true;
            break;
          }
        }
        if (tf) break;
        l++;
        d[a[l]]--;
        if (d[a[l]] == 0) c--;
      }
      int m = 0;
      if (c == k) {
        for (int i = 0; i <= 9; i++)
          if (d[i] != 0) {
            m = i;
            break;
          }
      }
      for (int i = l - 1; i >= 0; i--) b[i] = m;
    }
    for (int i = bl - 1; i >= 0; i--) printf("%d", b[i]);
    printf("\n");
  }
  return 0;
}

```