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;
int n;
void clean() {}
void solve() {
  clean();
  int a = 0, b = 1023;
  for (int x, i = 1; i <= n; i++) {
    char s[5];
    scanf("%s %d", s, &x);
    switch (s[0]) {
      case '^':
        a ^= x, b ^= x;
        break;
      case '|':
        a |= x, b |= x;
        break;
      case '&':
        a &= x, b &= x;
        break;
    }
  }
  int A = 0, X = 0, O = 0, bs = 1;
  for (int i = 0; i < 10; i++) {
    int taki = a & 1, mitsuha = b & 1;
    if (taki == 0) {
      if (mitsuha == 0) {
      }
      if (mitsuha == 1) A += bs;
    } else if (taki == 1) {
      if (mitsuha == 0) X += bs;
      if (mitsuha == 1) A += bs, O += bs;
    }
    a >>= 1, b >>= 1, bs <<= 1;
  }
  printf("3\n| %d\n^ %d\n& %d\n", O, X, A);
}
int main() {
  scanf("%d", &n), solve();
  return 0;
}
