Codeforces Round #443 (Div. 2): C. Short Program

C. Short Program

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.

In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.

Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya's program, and consists of no more than 5 lines. Your program should return the same integer as Petya's program for all arguments from 0 to 1023.

Input

The first line contains an integer n (1 ≤ n ≤ 5·105) — the number of lines.

Next n lines contain commands. A command consists of a character that represents the operation ("&", "|" or "^" for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.

Output

Output an integer k (0 ≤ k ≤ 5) — the length of your program.

Next k lines must contain commands in the same format as in the input.

Examples

input

3| 3^ 2| 1

output

2| 3^ 2

input

3& 1& 3& 5

output

1& 1

input

3^ 1^ 2^ 3

output

0

题意：

给你一个只包含('&', '^', '|')的计算步骤（输入及数据范围[0, 1023]），你需要构造出另一个计算步骤，要求

①两个计算步骤等效，即同样的输入能得出同样的输出

③你构造的计算步骤不能超过5步

结论：只需要一次^，一次&，一次|即可

方法：令x = 0，y = 1023，之后通过题目中的计算步骤算出x'和y'

通过判断x'和y'每一位的状态判断当前位是^还是&还是|

复杂度O(n)

#include<stdio.h>int a[500005];char str[500005];int main(void){int n, i, ans1, ans2, ans, c, d, e;scanf("%d", &n);for(i=1;i<=n;i++)scanf(" %c%d", &str[i], &a[i]);ans1 = 0, ans2 = 1023;for(i=1;i<=n;i++){if(str[i]=='|')ans1 |= a[i], ans2 |= a[i];else if(str[i]=='&')ans1 &= a[i], ans2 &= a[i];elseans1 ^= a[i], ans2 ^= a[i];}c = e = 0;d = 1023;for(i=0;i<=9;i++){if(ans1&(1<<i) && ans2&(1<<i))c ^= (1<<i);else if(ans1&(1<<i) && (ans2&(1<<i))==0)e ^= (1<<i);else if((ans1&(1<<i))==0 && (ans2&(1<<i))==0)d ^= (1<<i);}printf("3\n");printf("& %d\n", d);printf("^ %d\n", e);printf("| %d\n", c);return 0;}