CodeForces

 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 x_i 0 ≤ x_i ≤ 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.

Example

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 inhttps://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.

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题意：用五次运算以内表示题目给的运算。就是0到1023的所有数经过你的运算和题目给的运算结果必须一样，而且在5次以内。

思路：用 &，^ , |  表示题目给的运算，所以三步就可以，就变成   和 几 进行这几个运算的问题。所以就假设x和y。x=1023(10),1111111111(2).y=0(10),0000000000(2);  分别把x和y经过题目的运算，再来看xy每一位（二进制）的变化。刚开始x每一位（二进制）是1,y的每一位（二进制）是0，当运算结束，x和y的每一位（二进制）结果可以用来推算出这一位（二进制）分别经历了什么（2333）。

比如：

x1(表示二进制 x 的第一位数） ,y1(表示二进制 y 的第一位数）

开始 x1 = 1; y1 = 0;

结果 x1 = 0; y1 = 1;

那么 解方程  1 & （？）^（？）|（？） = 0；  0 & （？）^（？）|（？） = 1；

可以得出括号里依次为  1 ， 1 ， 0；

（& :与： 1&1=1，1&0=0&1=1，0&0=0）

（^ ：异或：1^1=0^0=0  ,  1^0=0^1=1 ）

（| ：或： 1|1=0|1=1|0=1，0|0=0）

#include<map>
#include<set>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f
#define ll long long
#define maxn 13005
using namespace std;
int x=1023,y=0,And,Or,Xor;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        char a;int m;
        cin>>a>>m;
        if(a=='&'){
            x&=m;y&=m;
        }
        if(a=='|'){
            x|=m;y|=m;
        }
        if(a=='^'){
            x^=m;y^=m;
        }
    }
    int res=1,xx,yy;
    for(int i=0;i<10;i++)
    {
        xx= (x&1);
        yy= (y&1);
        if(xx==1){
            if(yy==1){
                And+=res;
                Xor+=0;
                Or+=res;
            }
            else{
                And+=res;
                Xor+=0;
                Or+=0;
            }
        }
        else{
            if(yy==1){
                And+=res;
                Xor+=res;
                Or+=0;
            }
            else {
                And+=0;
                Xor+=0;
                Or+=0;
            }
        }
        x>>=1;y>>=1;
        res<<=1;
    }
    printf("3
& %d
^ %d
| %d
",And,Xor,Or);
}