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.

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 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.

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.

思路：取x=0,y=1023（即两个互为取反数的数字）。将n次位运算操作后的x，y值，对比即可得到每一位数字进行的 操作。

代码：

 #include<bits/stdc++.h>

 #define db double

 #include<vector>

 #define ll long long

 #define vec vector<ll>

 #define Mt  vector<vec>

 #define ci(x) scanf("%d",&x)

 #define cd(x) scanf("%lf",&x)

 #define cl(x) scanf("%lld",&x)

 #define pi(x) printf("%d\n",x)

 #define pd(x) printf("%f\n",x)

 #define pl(x) printf("%lld\n",x)

 const int N = 1e6 + ;

 const int mod = 1e9 + ;

 const int MOD = mod - ;

 const db  eps = 1e-;

 const db  PI = acos(-1.0);

 using namespace std;

 bool cal(int a,int i){

     if(a&(<<i)) return ;

     return ;

 }

 int main(){

     int n;ci(n);

     int x=,y=;

     while (n--) {

         char c[];

         int t;

         scanf("%s %d", c, &t);

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

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

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

     }

     /*

      4种:

      x:0,y:1.

      x y:0011,0010,0111,0110

      */

     int v1 = , v2 = , v3 = ;

     for (int i = ; i < ; i++) {

         if(!cal(x,i)&&cal(y,i))  v2+=(<<i);//0011: |0 &1 ^0

         if(cal(x,i)&&cal(y,i))   v3+=(<<i);//0111: |0 &0 ^1

         if(cal(x,i)&&!cal(y,i))  v2+=(<<i),v3+=(<<i);//0110: |0 &1 ^1

     }

     printf("3\n");

     printf("| %d\n", v1);

     printf("& %d\n", v2);

     printf("^ %d\n", v3);

     return ;

 }