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>
using namespace std;
const int maxn = 5e5 + 100;
int op[maxn], arr[maxn];
int n;
int res[3][10];
void solve() {
  for (int i = 0; i < 10; i++) res[0][i] = 1;
  int a = (1 << 10) - 1, b = 0;
  for (int i = 0; i < n; i++) {
    if (op[i] == 0)
      a &= arr[i], b &= arr[i];
    else if (op[i] == 1)
      a |= arr[i], b |= arr[i];
    else
      a ^= arr[i], b ^= arr[i];
  }
  for (int i = 0; i < 10; i++) {
    int x = (a >> i) & 1, y = (b >> i) & 1;
    if (!x && !y)
      res[0][i] = 0;
    else if (!x && y)
      res[2][i] = 1;
    else if (x && y)
      res[1][i] = 1;
  }
  int ans[3];
  memset(ans, 0, sizeof(ans));
  for (int i = 0; i < 10; i++)
    for (int j = 0; j < 3; j++) ans[j] += (1 << i) * res[j][i];
  printf("3\n& %d\n| %d\n^ %d\n", ans[0], ans[1], ans[2]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    char s[5];
    scanf("%s", s);
    if (s[0] == '&')
      op[i] = 0;
    else if (s[0] == '|')
      op[i] = 1;
    else
      op[i] = 2;
    scanf("%d", &arr[i]);
  }
  solve();
  return 0;
}
