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;
const int maxn = 5e5 + 10;
int aa[10][2], num[10][2], tt[maxn];
char s[maxn][5];
int main() {
  int n;
  scanf("%d", &n);
  for (int j = 0; j < 10; j++) {
    aa[j][0] = 0;
    aa[j][1] = 1;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%s%d", s[i], &tt[i]);
    for (int j = 0; j < 10; j++) {
      if (s[i][0] == '|') {
        aa[j][0] = ((tt[i] >> j) & 1) | aa[j][0];
        aa[j][1] = ((tt[i] >> j) & 1) | aa[j][1];
      } else if (s[i][0] == '^') {
        aa[j][0] = ((tt[i] >> j) & 1) ^ aa[j][0];
        aa[j][1] = ((tt[i] >> j) & 1) ^ aa[j][1];
      } else {
        aa[j][0] = ((tt[i] >> j) & 1) & aa[j][0];
        aa[j][1] = ((tt[i] >> j) & 1) & aa[j][1];
      }
    }
  }
  for (int i = 0; i <= 1023; i++) {
    for (int j = 0; j <= 1023; j++) {
      for (int k = 0; k < 10; k++) {
        num[k][0] = 0;
        num[k][1] = 1;
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((i >> k) & 1) | num[k][0];
        num[k][1] = ((i >> k) & 1) | num[k][1];
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((j >> k) & 1) ^ num[k][0];
        num[k][1] = ((j >> k) & 1) ^ num[k][1];
      }
      int flag = 1;
      for (int k = 0; k < 10; k++) {
        if (aa[k][0] != num[k][0] || aa[k][1] != num[k][1]) {
          flag = 0;
          break;
        }
      }
      if (flag) {
        printf("2\n| %d\n^ %d\n", i, j);
        return 0;
      }
    }
  }
  for (int i = 0; i <= 1023; i++) {
    for (int j = 0; j <= 1023; j++) {
      for (int k = 0; k < 10; k++) {
        num[k][0] = 0;
        num[k][1] = 1;
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((i >> k) & 1) | num[k][0];
        num[k][1] = ((i >> k) & 1) | num[k][1];
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((j >> k) & 1) & num[k][0];
        num[k][1] = ((j >> k) & 1) & num[k][1];
      }
      int flag = 1;
      for (int k = 0; k < 10; k++) {
        if (aa[k][0] != num[k][0] || aa[k][1] != num[k][1]) {
          flag = 0;
          break;
        }
      }
      if (flag) {
        printf("| %d\n& %d\n", i, j);
        return 0;
      }
    }
  }
  for (int i = 0; i <= 1023; i++) {
    for (int j = 0; j <= 1023; j++) {
      for (int k = 0; k < 10; k++) {
        num[k][0] = 0;
        num[k][1] = 1;
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((i >> k) & 1) ^ num[k][0];
        num[k][1] = ((i >> k) & 1) ^ num[k][1];
      }
      for (int k = 0; k < 10; k++) {
        num[k][0] = ((j >> k) & 1) & num[k][0];
        num[k][1] = ((j >> k) & 1) & num[k][1];
      }
      int flag = 1;
      for (int k = 0; k < 10; k++) {
        if (aa[k][0] != num[k][0] || aa[k][1] != num[k][1]) {
          flag = 0;
          break;
        }
      }
      if (flag) {
        printf("^ %d\n& %d\n", i, j);
        return 0;
      }
    }
  }
}