Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
const int N = 1000000;
int n, v[N];
int x, y = (1023), z;
char ch[N], str[N];
inline int getans(int BIT, int val) {
  int now = val;
  for (int i = 1; i <= n; i++) {
    if (ch[i] == '|') now = now | (v[i] >> BIT);
    if (ch[i] == '&') now = now & (v[i] >> BIT);
    if (ch[i] == '^') now = now ^ (v[i] >> BIT);
  }
  return now & 1;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", str);
    ch[i] = str[0];
    read(v[i]);
  }
  for (int i = 0; i <= 9; i++) {
    int v0 = getans(i, 0), v1 = getans(i, 1);
    if (v0 == 1 && v1 == 0)
      x |= (1 << i);
    else if (v0 == 0 && v1 == 0)
      y ^= (1 << i);
    else if (v0 == 1 && v1 == 1)
      z |= (1 << i);
  }
  puts("3");
  printf("^ %d\n", x);
  printf("& %d\n", y);
  printf("| %d\n", z);
  return 0;
}