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 N = 1e5 + 5;
pair<char, int> a[N];
int n;
int res[12][5];
int ans(int num) {
  int res = num;
  for (int i = 1; i <= n; ++i)
    if (a[i].first == '|')
      res = res | a[i].second;
    else if (a[i].first == '&')
      res = res & a[i].second;
    else
      res = res ^ a[i].second;
  return res;
}
int sum(int z) {
  int s = 0;
  for (int i = 0; i <= 9; i++) s += res[i][z] << i;
  return s;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf(" %c%d", &a[i].first, &a[i].second);
  for (int z = 9; z >= 0; z--) {
    int u = ans(0 << z) >> z;
    int v = ans(1 << z) >> z;
    if (u == 1 && v == 1)
      res[z][3] = 1;
    else if (u == 1 && v == 0)
      res[z][2] = 1;
    else if (u == 0 && v == 1)
      res[z][1] = res[z][2] = 1;
  }
  printf("3\n");
  printf("^ %d\n", sum(1));
  printf("& %d\n", sum(2));
  printf("| %d\n", sum(3));
  return 0;
}
