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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
int a = 0, b = 0, c = 0;
int res[10][2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (0), i_end_ = (10); i < i_end_; ++i)
    for (int j = (0), j_end_ = (2); j < j_end_; ++j) res[i][j] = j;
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) {
    static char s[5];
    scanf("%s", s);
    int first;
    scanf("%d", &first);
    for (int j = (0), j_end_ = (10); j < j_end_; ++j)
      for (int k = (0), k_end_ = (2); k < k_end_; ++k) {
        if (s[0] == '&') res[j][k] &= (first >> j & 1);
        if (s[0] == '|') res[j][k] |= (first >> j & 1);
        if (s[0] == '^') res[j][k] ^= (first >> j & 1);
      }
  }
  for (int i = (0), i_end_ = (10); i < i_end_; ++i) {
    if (res[i][0] != res[i][1]) {
      a ^= 1 << i;
      if (!res[i][0]) c ^= 1 << i;
    } else {
      b ^= 1 << i;
      if (!res[i][0]) c ^= 1 << i;
    }
  }
  printf("3\n");
  printf("^ %d\n", a);
  printf("| %d\n", b);
  printf("^ %d\n", c);
  return 0;
}
