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 XOR, OR;
int p[11];
int b1, b2;
char ch[500005];
int num[500005][11];
int cal(int b, int j, int i) {
  if (ch[j] == '|') return b | num[j][i];
  if (ch[j] == '&')
    return b & num[j][i];
  else
    return b ^ num[j][i];
}
int main() {
  int n, m;
  p[0] = 1;
  for (int i = 1; i < 11; i++) p[i] = p[i - 1] * 2;
  while (cin >> n) {
    XOR = OR = 0;
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; i++) {
      scanf(" %c %d", &ch[i], &m);
      int t = 0;
      while (m) {
        if (m & 1) num[i][t] = 1;
        t++;
        m >>= 1;
      }
    }
    for (int i = 0; i < 10; i++) {
      b1 = 1, b2 = 0;
      for (int j = 0; j < n; j++) {
        b1 = cal(b1, j, i);
        b2 = cal(b2, j, i);
      }
      if (b1 == 0 && b2 == 0) {
        OR ^= p[i];
        XOR ^= p[i];
      }
      if (b1 == 1 && b2 == 1) OR ^= p[i];
      if (b1 == 0 && b2 == 1) XOR ^= p[i];
    }
    int sum = 0;
    if (OR) sum++;
    if (XOR) sum++;
    cout << sum << endl;
    if (OR) cout << '|' << ' ' << OR << endl;
    if (XOR) cout << '^' << ' ' << XOR << endl;
  }
}