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 inf = 1e9 + 5;
const double eps = 0;
const int ms = 0;
const int md = 0;
const int mod = 1e9 + 9;
const int pi = 31, pii = 41;
const int base = 1e9;
const int N = 5e5 + 10;
const int NN = 1e5 + 5;
const int M = 1e5 + 5;
const int blocks = 600;
void readfile() {
  freopen("textin.inp", "r", stdin);
  freopen("text.out", "w", stdout);
}
int n;
pair<char, int> a[N];
int an = 1023, o = 0, x = 0;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  int cnt = 0;
  for (int i = 1; i <= n; i++) cin >> a[i].first >> a[i].second;
  for (int i = 0; i <= 9; i++) {
    int set = 1, unset = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j].first == '&') {
        set &= ((a[j].second >> i) & 1);
        unset &= ((a[j].second >> i) & 1);
      } else if (a[j].first == '|') {
        set |= ((a[j].second >> i) & 1);
        unset |= ((a[j].second >> i) & 1);
      } else {
        set ^= ((a[j].second >> i) & 1);
        unset ^= ((a[j].second >> i) & 1);
      }
    }
    if (set == unset && set == 1) {
      o += (1 << i);
    } else if (set == unset && set == 0) {
      an -= (1 << i);
    } else {
      if (set == 0 && unset == 1) {
        x += (1 << i);
      }
    }
  }
  cout << 3 << '\n';
  cout << "& " << an << '\n';
  cout << "| " << o << '\n';
  cout << "^ " << x << '\n';
  return 0;
}
