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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
int n, x[500000 + 5] = {0};
int a[500005][11] = {0}, ans1[11] = {0};
char ch[500000 + 5], ans2[11] = {};
int f(int i, int tmp) {
  for (int j = 0; j < n; j++) {
    if (ch[j] == '&') tmp = tmp & a[j][i];
    if (ch[j] == '|') tmp = tmp | a[j][i];
    if (ch[j] == '^') tmp = tmp ^ a[j][i];
  }
  return tmp;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%c", &ch[i]);
    scanf("%c %d", &ch[i], &x[i]);
    for (int j = 0; j < 10; j++) {
      a[i][j] = x[i] & 1;
      x[i] /= 2;
      if (!x[i]) break;
    }
  }
  for (int i = 0; i < 10; i++) {
    int a1 = f(i, 0), b1 = f(i, 1);
    if (a1 == 0 && b1 == 0) {
      ans1[i] = 0;
      ans2[i] = '&';
    } else if (a1 == 0 && b1 == 1) {
      ans1[i] = 0;
      ans2[i] = '^';
    } else if (a1 == 1 && b1 == 0) {
      ans1[i] = 1;
      ans2[i] = '^';
    } else if (a1 == 1 && b1 == 1) {
      ans1[i] = 1;
      ans2[i] = '|';
    }
  }
  printf("3\n");
  int ans = 1;
  for (int i = 9; i >= 0; i--) {
    ans = ans * 2 + 1;
    if (ans2[i] == '&') ans--;
  }
  printf("& %d\n", ans - 1024);
  ans = 0;
  for (int i = 9; i >= 0; i--) {
    ans *= 2;
    if (ans2[i] == '^') ans += ans1[i];
  }
  printf("^ %d\n", ans);
  ans = 0;
  for (int i = 9; i >= 0; i--) {
    ans *= 2;
    if (ans2[i] == '|') ans += ans1[i];
  }
  printf("| %d\n", ans);
  return 0;
}