Codeforces Round #443 (Div. 2)

C. Short Program

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

Note

You can read about bitwise operations in https://en.wikipedia.org/wiki/Bitwise_operation.

Second sample:

Let x be an input of the Petya's program. It's output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.

题意：给一个任意数x，进行位运算，求怎么简化到不超过5次。

分析：看出每次的操作数不超过2^10-1，先用0000000000,1111111111，进行题意的操作，发现规律，01-----> 0/1，通过 | ^ & 运算使得它成立。

#include <bits/stdc++.h>

using namespace std;

bool calc(int a,int i) {
    if(a&(1<<i)) return 1;
    return 0;
}

int main()
{
    int n;
    int x = 0,y = 1023;

    cin>>n;
    while(n--) {
        char str[5];
        int t;
        scanf("%s%d",str,&t);

        if(str[0]=='|') x|=t,y|=t;

        if(str[0]=='&') x&=t,y&=t;

        if(str[0]=='^') x^=t,y^=t;

    }

    int v1 = 0; // |
    int v2 = 0; // ^
    int v3 = 1023;
    for(int i = 0; i < 10; i++) {
        if(calc(x,i)&&calc(y,i)) v1 = v1 + (1<<i);
        if(calc(x,i)&&!calc(y,i)) v2 = v2 + (1<<i);
        //if(!calc(x,i)&&calc(y,i))
        if(!calc(x,i)&&!calc(y,i)) v3 = v3 - (1<<i);
    }


    printf("3
");

    printf("| %d
",v1);
    printf("^ %d
",v2);
    printf("& %d
",v3);


    return 0;
}