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;
int n;
char opt[500010][3];
int x[500010], ans[500010][2];
int getBit(int u, int x) { return x >> u & 1; }
int calc(char u, int a, int b) {
  if (u == '|') return a | b;
  if (u == '&') return a & b;
  return a ^ b;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s%d", opt[i], &x[i]);
  }
  for (int i = 0; i < 10; ++i) {
    for (int t = 0; t < 2; ++t) {
      ans[i][t] = t;
      for (int j = 1; j <= n; ++j) {
        ans[i][t] = calc(opt[j][0], ans[i][t], getBit(i, x[j]));
      }
    }
  }
  int ans1 = 0, ans2 = 0, ans3 = 0;
  for (int i = 9; i >= 0; --i) {
    if (!ans[i][0] && !ans[i][1]) {
      ans1 = ans1 << 1 | 1;
      ans2 = ans2 << 1 | 1;
      ans3 = ans3 << 1 | 0;
    } else if (ans[i][0] && ans[i][1]) {
      ans1 = ans1 << 1 | 1;
      ans2 = ans2 << 1 | 0;
      ans3 = ans3 << 1 | 1;
    } else if (ans[i][0] && !ans[i][1]) {
      ans1 = ans1 << 1 | 0;
      ans2 = ans2 << 1 | 1;
      ans3 = ans3 << 1 | 1;
    } else if (!ans[i][0] && ans[i][1]) {
      ans1 = ans1 << 1 | 0;
      ans2 = ans2 << 1 | 0;
      ans3 = ans3 << 1 | 1;
    }
  }
  printf("3\n| %d\n^ %d\n& %d\n", ans1, ans2, ans3);
  return 0;
}