Codeforces Round #443 (Div. 2)

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

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：

给你n个位运算，假设x属于[0,1023]，让你求出特定的0-5个位运算，使这两种位运算后 答案相同。

POINT：

观察x属于[0，1023]这个条件，我们用10位全是1的1023和10位全是0的0，分别对给我们的n个位运算运算后。

得到2个10长度1 0串的。然后根据两个串的1，0变化成1或0来确定&^|运算。

所以答案必定是3个。

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <queue>using namespace std;#define LL long longconst LL inf = 0x3f3f3f3f;const int maxn = 5e5+23;char s[maxn];int x[maxn];int n;int a1[13];int a2[13];void doit1(int now){    for(int i=1;i<=n;i++){        if(s[i]=='|') now=now|x[i];        if(s[i]=='^') now=now^x[i];        if(s[i]=='&') now=now&x[i];    }    int k=1;    while(k!=11){        a1[k++]=now&1;        now>>=1;    }    }void doit2(int now){    for(int i=1;i<=n;i++){        if(s[i]=='|') now=now|x[i];        if(s[i]=='^') now=now^x[i];        if(s[i]=='&') now=now&x[i];    }    int k=1;    while(k!=11){        a2[k++]=now&1;        now>>=1;    }    }int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++){        cin>>s[i]>>x[i];    }    doit1(1023);    doit2(0);    int b1=0,b2=0,b3=0;//& ^ |//    for(int i=10;i>=1;i--){//        printf("%d ",a1[i]);//    }//    printf("\n");//    for(int i=10;i>=1;i--){//        printf("%d ",a2[i]);//    }//    printf("\n");    b1=1023;    for(int i=10;i>=1;i--){        if(a1[i]==1&&a2[i]==1){            b3=b3+(1<<(i-1));        }        if(a1[i]==0&&a2[i]==0){            b1=b1-(1<<(i-1));        }        if(a1[i]==0&&a2[i]==1){            b2=b2+(1<<(i-1));        }    }        printf("3\n");    printf("^ %d\n| %d\n& %d\n",b2,b3,b1);    }