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;
long long mod = 1e9 + 7;
long long modexp(long long x, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long y = modexp(x, n / 2) % mod;
    return (y * y) % mod;
  }
  return (x * modexp(x, n - 1) % mod) % mod;
}
vector<pair<char, int> > v;
int getans(char ch, int x, int y) {
  if (ch == '|')
    return x | y;
  else if (ch == '&')
    return x & y;
  else
    return x ^ y;
}
int xor1[10];
int xor2[10];
int main() {
  int n, x, y;
  scanf("%d", &n);
  char ch;
  int bits[10];
  for (int i = 0; i < 10; i++) bits[i] = 5;
  char prev = '$';
  for (int i = 0; i < n; i++) {
    cin >> ch;
    scanf("%d", &x);
    v.push_back(make_pair(ch, x));
  }
  for (int i = 0; i < v.size(); i++) {
    int bin[10];
    int j = 9;
    int temp = v[i].second;
    ch = v[i].first;
    for (int k = 0; k < 10; k++) bin[k] = 0;
    while (temp) {
      bin[j--] = temp % 2;
      temp /= 2;
    }
    if (ch == '|') {
      for (j = 0; j < 10; j++) {
        if (bin[j] == 1) {
          bits[j] = 1;
          xor1[j] = 0;
        }
      }
    } else if (ch == '&') {
      for (j = 0; j < 10; j++) {
        if (bin[j] == 0) {
          bits[j] = 0;
          xor1[j] = 0;
        }
      }
    } else {
      for (j = 0; j < 10; j++) {
        if (bin[j] == 1) {
          if (bits[j] == 1) {
            bits[j] = 0;
          } else if (bits[j] == 0) {
            bits[j] = 1;
          } else {
            xor1[j]++;
          }
        }
      }
    }
  }
  int ans[10];
  printf("3\n");
  for (int i = 0; i < 10; i++) {
    if (bits[i] == 0) {
      ans[i] = 0;
    } else if (bits[i] == 1)
      ans[i] = 1;
    else
      ans[i] = 0;
  }
  int ans1 = 0;
  for (int i = 0; i < 10; i++) {
    ans1 += ans[i] * pow(2, 10 - i - 1);
  }
  printf("| %d\n", ans1);
  for (int i = 0; i < 10; i++) {
    if (bits[i] == 0) {
      ans[i] = 0;
    } else
      ans[i] = 1;
  }
  ans1 = 0;
  for (int i = 0; i < 10; i++) {
    ans1 += ans[i] * pow(2, 10 - i - 1);
  }
  printf("& %d\n", ans1);
  for (int i = 0; i < 10; i++) {
    if (xor1[i] % 2) {
      ans[i] = 1;
    } else
      ans[i] = 0;
  }
  ans1 = 0;
  for (int i = 0; i < 10; i++) {
    ans1 += ans[i] * pow(2, 10 - i - 1);
  }
  printf("^ %d", ans1);
  return 0;
}