Problem: 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
&amp; 1
&amp; 3
&amp; 5


Output

1
&amp; 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, a[500050], x[12];
char b[500050];
vector<int> Xor, Or, And;
int trans(int num, char opp, int bnum) {
  if (num == 0) {
    if (opp == '|') {
      return 0 | bnum;
    } else if (opp == '^') {
      return 0 ^ bnum;
    } else {
      return 0 & bnum;
    }
  } else if (num == 1) {
    if (opp == '|') {
      return 1 | bnum;
    } else if (opp == '^') {
      return 1 ^ bnum;
    } else {
      return 1 & bnum;
    }
  } else if (num == 2) {
    if (opp == '|') {
      if (bnum == 1) {
        return 1;
      } else {
        return num;
      }
    } else if (opp == '^') {
      if (bnum == 1) {
        return 3;
      } else {
        return 2;
      }
    } else {
      if (bnum == 1) {
        return 2;
      } else {
        return 0;
      }
    }
  } else {
    if (opp == '|') {
      if (bnum == 1) {
        return 1;
      } else {
        return 3;
      }
    } else if (opp == '^') {
      if (bnum == 1) {
        return 2;
      } else {
        return 3;
      }
    } else {
      if (bnum == 1) {
        return 3;
      } else {
        return 0;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 12; i++) x[i] = 2;
  for (int i = 0; i < n; i++) {
    getchar();
    scanf("%c %d", &b[i], &a[i]);
    for (int j = 0; j < 12; j++) {
      int y = (1 << j) & a[i];
      if (y) y = 1;
      x[j] = trans(x[j], b[i], y);
    }
  }
  for (int i = 0; i < 10; i++) {
    if (x[i] == 1)
      Or.push_back(i);
    else if (x[i] == 0)
      And.push_back(i);
    else if (x[i] == 3)
      Xor.push_back(i);
  }
  printf("%d\n", 3);
  printf("| ");
  int curr = 0;
  for (auto next : Or) {
    curr += (1 << next);
  }
  printf("%d\n", curr);
  printf("& ");
  curr = 1023;
  for (auto next : And) {
    curr -= (1 << next);
  }
  printf("%d\n", curr);
  printf("^ ");
  curr = 0;
  for (auto next : Xor) {
    curr += (1 << next);
  }
  printf("%d\n", curr);
  return 0;
}