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;
int ans1[15];
int ans0[15];
std::vector<pair<char, int> > v;
int op1[15];
int op2[15];
int op3[15];
void makeans(char op, int nums) {
  int tmp;
  for (int i = 0; i < 10; i++) {
    tmp = (nums & 1);
    if (op == '|') {
      ans1[i] |= tmp;
      ans0[i] |= tmp;
    }
    if (op == '^') {
      ans0[i] ^= tmp;
      ans1[i] ^= tmp;
    }
    if (op == '&') {
      ans1[i] &= tmp;
      ans0[i] &= tmp;
    }
    nums = (nums >> 1);
  }
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) ans1[i] = 1, op3[i] = 1;
  for (int i = 0; i < n; i++) {
    char op;
    int nums;
    scanf(" %c%d", &op, &nums);
    v.push_back(make_pair(op, nums));
    makeans(op, nums);
  }
  int ans = 0;
  int flag[5] = {0, 0, 0, 0};
  for (int i = 0; i < 10; i++) {
    if (ans0[i] == 0 && ans1[i] == 1) continue;
    if (ans0[i] == 1 && ans1[i] == 1) {
      op1[i] = 1;
      if (!flag[1]) ans++;
      flag[1] = 1;
    }
    if (ans0[i] == 0 && ans1[i] == 0) {
      op3[i] = 0;
      if (!flag[2]) ans++;
      flag[2] = 2;
    }
    if (ans0[i] == 1 && ans1[i] == 0) {
      op2[i] = 1;
      if (!flag[3]) ans++;
      flag[3] = 3;
    }
  }
  int num = 0;
  if (ans > n) {
    printf("%d\n", n);
    for (int i = 0; i < v.size(); i++) {
      printf("%c %d\n", v[i].first, v[i].second);
    }
    return 0;
  } else if (ans >= 3) {
    for (int i = 0; i < 10; i++) {
      if (op3[i] == 0) {
        op1[i] = 1;
        op2[i] = 1;
        op3[i] = 1;
      }
    }
    ans--;
  }
  printf("%d\n", ans);
  num = 0;
  for (int i = 9; i >= 0; i--) {
    num = (num << 1) + op1[i];
  }
  if (num != 0) printf("| %d\n", num);
  num = 0;
  for (int i = 9; i >= 0; i--) {
    num = (num << 1) + op2[i];
  }
  if (num != 0) printf("^ %d\n", num);
  num = 0;
  for (int i = 9; i >= 0; i--) {
    num = (num << 1) + op3[i];
  }
  if (num != 1023) printf("& %d\n", num);
  return 0;
}