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>
using namespace std;
const int MAXN = 500005;
int a[MAXN];
int opt[MAXN], dt[MAXN];
int n;
char s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("\n%c %d", &s, &dt[i]);
    if (s == '|')
      opt[i] = 0;
    else if (s == '&')
      opt[i] = 1;
    else
      opt[i] = 2;
  }
  int land = (1 << 10) - 1, lxor = 0, lor = 0;
  for (int i = 0; i < 10; i++) {
    int zero = 0, one = 1;
    for (int j = 1; j <= n; j++) {
      if (opt[j] == 0)
        zero |= ((dt[j] >> i) & 1), one |= ((dt[j] >> i) & 1);
      else if (opt[j] == 1)
        one &= ((dt[j] >> i) & 1), zero &= ((dt[j] >> i) & 1);
      else {
        one ^= ((dt[j] >> i) & 1);
        zero ^= ((dt[j] >> i) & 1);
      }
    }
    if (zero == 0 && one == 1)
      ;
    else if (zero == 1 && one == 1)
      lor |= (1 << i);
    else if (zero == 0 && one == 0)
      land ^= (1 << i);
    else
      lxor |= (1 << i);
  }
  cout << 3 << endl;
  cout << "& " << land << endl;
  cout << "| " << lor << endl;
  cout << "^ " << lxor << endl;
  return 0;
}
