1
& 0
 1
^ 123
 10
^ 218
& 150
| 935
& 61
| 588
& 897
| 411
| 584
^ 800
| 704
 3
& 1
& 3
& 5
 1
| 123
 1
^ 0
 1
| 1023
 3
& 242
^ 506
^ 522
 10
^ 160
& 1021
& 510
^ 470
& 1022
& 251
& 760
& 1016
| 772
| 515
 1
& 1023
 1
| 0
 2
| 56
^ 875
 1
^ 1023
 2
| 999
^ 689
 3
^ 125
^ 377
& 1019
 1
& 123

3
& 0
| 0
^ 0
 3
& 1023
| 0
^ 123
 3
& 763
| 763
^ 0
 3
& 1
| 0
^ 0
 3
& 1023
| 123
^ 0
 3
& 1023
| 0
^ 0
 3
& 1023
| 1023
^ 0
 3
& 1010
| 768
^ 240
 2
| 775
^ 112
 3
& 1023
| 0
^ 0
 3
& 1023
| 0
^ 0
 3
& 983
| 16
^ 835
 3
& 1023
| 0
^ 1023
 3
& 350
| 326
^ 16
 3
& 1019
| 0
^ 256
 3
& 123
| 0
^ 0

#include <bits/stdc++.h>
int stats[11], num, n;
char S[2];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s %d", S, &num);
    for (int j = 0, k = 1; k < 1024; k <<= 1, j++) {
      if (S[0] == '|' && k & num)
        stats[j] = 2;
      else if (S[0] == '&' && !(k & num))
        stats[j] = 3;
      else if (S[0] == '^' && k & num)
        stats[j] ^= 1;
    }
  }
  int orA = 0, xorA = 0, andA = 1023;
  for (int i = 0, k = 1; i < 10; k <<= 1, i++) {
    if (stats[i] == 1)
      xorA |= k;
    else if (stats[i] == 2)
      orA |= k;
    else if (stats[i] == 3)
      andA ^= k;
  }
  printf("3\n| %d\n& %d\n^ %d\n", orA, andA, xorA);
}
