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.
#include <bits/stdc++.h>
void sf(int &x) { scanf("%d", &x); }
void sf(char &x) { scanf("%c", &x); }
void sf(double &x) { scanf("%lf", &x); }
void sf(char *x) { scanf("%s", x); }
void sf(long long &x) { scanf("%I64d", &x); }
void pf(int x) { printf("%d", x); }
void pf(char x) { printf("%c", x); }
void pf(double x) { printf("%f", x); }
void pf(char *x) { printf("%s", x); }
void pf(long long x) { printf("%I64d", x); }
void pf() { printf("\n"); }
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
using namespace std;
const double eps = 1e-6;
const long long mod = 1e9 + 7;
const int maxn = 500100;
const int maxe = 500100;
void Solve() {
  int n;
  scanf("%d", &n);
  int x1 = 0, x2 = 1023, num;
  char ch[2];
  for (int i = 0; i < n; i++) {
    scanf("%s %d", ch, &num);
    if (ch[0] == '&') {
      x1 = x1 & num;
      x2 = x2 & num;
    }
    if (ch[0] == '|') {
      x1 = x1 | num;
      x2 = x2 | num;
    }
    if (ch[0] == '^') {
      x1 = x1 ^ num;
      x2 = x2 ^ num;
    }
  }
  int y1 = x1 & x2;
  int y2 = x1 | x2;
  int ans = 0;
  for (int i = 0; i < 10; i++) {
    if ((x1 >> i & 1) && !(x2 >> i & 1)) ans |= 1 << i;
  }
  printf("3\n");
  printf("| %d\n", y1);
  printf("& %d\n", y2);
  printf("^ %d\n", ans);
}
int main() {
  std::ios::sync_with_stdio(false);
  long long Case = 1, cases;
  while (Case--) {
    Solve();
  }
  return 0;
}
