先考虑每一位要变成 1 需要翻转的次数，从左往右考虑，每一位要变成 1 最多需要一次翻转。反过来想，每一位也最多需要一次固定操作，固定总操作数是 $k$，从左往右计算直到用完 $k$ 次，如果用不完，那么把剩下的次数都给最右边的位，因为它对字典序的影响最小。

// Date: Sun Dec 10 20:11:10 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, n, k, d[N];
string s;

int main(void) {
#ifdef _DEBUG
  freopen("1659b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n >> k >> s;

    int cur = k;
    memset(d, 0, sizeof d);

    For(i, 0, n) {
      if (!cur)
        break;

      if (s[i] == '1') {
        if (k & 1) {
          cur--;
          d[i] = 1;
        } else {
        }
      } else {
        if (k & 1) {
        } else {
          cur--;
          d[i] = 1;
        }
      }

      if (!cur)
        break;
    }

    if (cur) {
      d[n - 1] += cur;
    }

    For(i, 0, n) {
      int cnt = k - d[i];
      if (cnt & 1) {
        if (s[i] == '1')
          s[i] = '0';
        else
          s[i] = '1';
      }
    }

    cout << s << '\n';
    For(i, 0, n) { cout << d[i] << ' '; }
    cout << '\n';
  }

  return 0;
}