Output #1
2 1
1 2
2 1
* Initially $(S_1,S_2,S_3,S_4)=(\lbrace\rbrace,\lbrace\rbrace,\lbrace\rbrace,\lbrace\rbrace)$.
* The $1$\-st query adds $10$ to $S_1,S_2$, resulting in $(S_1,S_2,S_3,S_4)=(\lbrace 10\rbrace,\lbrace 10\rbrace,\lbrace\rbrace,\lbrace\rbrace)$.
* The $2$\-nd query adds $20$ to $S_2,S_3,S_4$, resulting in $(S_1,S_2,S_3,S_4)=(\lbrace 10\rbrace,\lbrace 10,20\rbrace,\lbrace 20\rbrace,\lbrace 20\rbrace)$.
* For the $3$\-rd query, the maximum number of elements among $S_1,S_2,S_3$ is $2$ achieved by $S_2$, so print `2 1`.
* The $4$\-th query removes $20$ from $S_1,S_2$, resulting in $(S_1,S_2,S_3,S_4)=(\lbrace 10\rbrace,\lbrace 10\rbrace,\lbrace 20\rbrace,\lbrace 20\rbrace)$.
* The $5$\-th query adds $10$ to $S_2,S_3$, resulting in $(S_1,S_2,S_3,S_4)=(\lbrace 10\rbrace,\lbrace 10\rbrace,\lbrace 10,20\rbrace,\lbrace 20\rbrace)$.
* For the $6$\-th query, the maximum number of elements among $S_1,S_2$ is $1$ achieved by $S_1,S_2$, so print `1 2`.
* For the $7$\-th query, the maximum number of elements among $S_1,S_2,S_3,S_4$ is $2$ achieved by $S_3$, so print `2 1`.