description: Alice and Borys are playing tennis.
  
  A tennis match consists of games. In each game, one of the players is serving and the other one is receiving.
  
  Players serve in turns: after a game where Alice is serving follows a game where Borys is serving, and vice versa.
  
  Each game ends with a victory of one of the players. If a game is won by the serving player, it's said that this player holds serve. If a game is won by the receiving player, it's said that this player breaks serve.
  
  It is known that Alice won a games and Borys won b games during the match. It is unknown who served first and who won which games.
  
  Find all values of k such that exactly k breaks could happen during the match between Alice and Borys in total.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.
  
  Each of the next t lines describes one test case and contains two integers a and b (0 ≤ a, b ≤ 10^5; a + b > 0) — the number of games won by Alice and Borys, respectively.
  
  It is guaranteed that the sum of a + b over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print two lines.
  
  In the first line, print a single integer m (1 ≤ m ≤ a + b + 1) — the number of values of k such that exactly k breaks could happen during the match.
  
  In the second line, print m distinct integers k_1, k_2, …, k_m (0 ≤ k_1 < k_2 < … < k_m ≤ a + b) — the sought values of k in increasing order.
  
  Example
  
  Input
  
  
  3
  2 1
  1 1
  0 5
  
  
  Output
  
  
  4
  0 1 2 3
  2
  0 2
  2
  2 3
  
  Note
  
  In the first test case, any number of breaks between 0 and 3 could happen during the match: 
  
    * Alice holds serve, Borys holds serve, Alice holds serve: 0 breaks; 
    * Borys holds serve, Alice holds serve, Alice breaks serve: 1 break; 
    * Borys breaks serve, Alice breaks serve, Alice holds serve: 2 breaks; 
    * Alice breaks serve, Borys breaks serve, Alice breaks serve: 3 breaks. 
  
  
  
  In the second test case, the players could either both hold serves (0 breaks) or both break serves (2 breaks).
  
  In the third test case, either 2 or 3 breaks could happen: 
  
    * Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve: 2 breaks; 
    * Borys breaks serve, Borys holds serve, Borys breaks serve, Borys holds serve, Borys breaks serve: 3 breaks. 
solution: #include <bits/stdc++.h>
  using namespace std;
  void debugs() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long multiply(long long x, long long res[], long long ressize) {
    long long carry = 0;
    for (long long i = 0; i < ressize; i++) {
      long long product = res[i] * x + carry;
      res[i] = product % 10;
      carry = product / 10;
    }
    while (carry) {
      res[ressize] = carry % 10;
      carry = carry / 10;
      ressize++;
    }
    return ressize;
  }
  void bigfactorial(long long n) {
    long long ressize = 1;
    long long res[10000];
    res[0] = 1;
    for (long long i = 2; i <= n; i++) {
      ressize = multiply(i, res, ressize);
    }
    for (long long i = ressize - 1; i >= 0; i--) {
      cout << res[i];
    }
  }
  bool substring(string s1, string s2) {
    int M = s1.length();
    int N = s2.length();
    for (int i = 0; i <= N - M; i++) {
      int j;
      for (j = 0; j < M; j++)
        if (s2[i + j] != s1[j]) break;
      if (j == M) return true;
      ;
    }
    return false;
    ;
  }
  long long n;
  bool palindrome(string s) {
    string t;
    t = s;
    reverse(t.begin(), t.end());
    if (s == t) {
      return true;
    }
    return false;
  }
  bool issafe(vector<vector<long long>> board, long long row, long long col) {
    for (long long i = 0; i < col; i++) {
      if (board[row][i]) return false;
    }
    for (long long i = row, j = col; i >= 0 && j >= 0; i--, j--) {
      if (board[i][j] == 1) return false;
    }
    for (long long i = row, j = col; i < n && j < n; i++, j++) {
      if (board[i][j] == 1) return false;
    }
    return true;
  }
  void display(vector<vector<long long>> &board) {
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) {
        cout << board[i][j] << " ";
      }
      cout << endl;
    }
  }
  bool solvenqueen(vector<vector<long long>> &board, long long col) {
    if (col >= n) {
      return true;
    }
    for (long long i = 0; i < n; i++) {
      if (issafe(board, i, col)) {
        board[i][col] = 1;
        if (solvenqueen(board, col + 1)) return true;
        board[i][col] = 0;
      }
    }
    return false;
  }
  int convertstring(string s) {
    stringstream geek(s);
    int x = 0;
    geek >> x;
    return x;
  }
  bool binary(long long k, long long a[]) {
    long long l = 1;
    long long r = 1e4;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (a[mid] == (k)) {
        return true;
      } else if (a[mid] > (k)) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    return false;
  }
  bool notprime(int n) {
    if (n == 2 || n == 3 || n == 5 || n == 7) {
      return false;
    }
    return true;
  }
  int bin(string n) {
    int num = convertstring(n);
    int dec_value = 0;
    int base = 1;
    int temp = num;
    while (temp) {
      int last_digit = temp % 10;
      temp = temp / 10;
      dec_value += last_digit * base;
      base = base * 2;
    }
    return dec_value;
  }
  bool sorted(int a[], int n) {
    for (int i = 1; i <= n - 1; i++) {
      if (a[i] > a[i + 1]) {
        return false;
      }
    }
    return true;
  }
  int main() {
    long long a, b;
    cin >> a >> b;
    int m = ((a + b) / 2) - min(a, b);
    int maxi = a + b - (((a + b) / 2) - min(a, b));
    if ((a + b) % 2 == 0) {
      cout << (maxi - m) / 2 + 1 << endl;
      for (int i = m; i <= maxi; i += 2) {
        cout << i << " ";
      }
      cout << endl;
    } else {
      cout << maxi - m + 1 << endl;
      for (int i = m; i <= maxi; i++) {
        cout << i << " ";
      }
      cout << endl;
    }
  }
  }
  