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 long long mod = 1000000007;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
template <class T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int maxn = 5e5 + 10;
int tb[10][2], ans[10][3], op[maxn], r[maxn], n;
int tran(char* ch) {
  if (ch[0] == '|') return 0;
  if (ch[0] == '&') return 1;
  return 2;
}
int cal(int x, int y, int o) {
  if (o == 0) return x | y;
  if (o == 1) return x & y;
  return x ^ y;
}
void get_tb() {
  for (int i = 0; i < 10; i++) tb[i][1] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 10; j++) {
      int y = (r[i] >> j) & 1;
      tb[j][0] = cal(tb[j][0], y, op[i]);
      tb[j][1] = cal(tb[j][1], y, op[i]);
    }
  }
  for (int i = 0; i < 10; i++) {
  }
}
void get_ans() {
  for (int i = 0; i < 10; i++) {
    ans[i][0] = 0;
    ans[i][1] = 1;
    ans[i][2] = 0;
  }
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 3; j++) {
      if (cal(0, 1, j) == tb[i][0] && cal(1, 1, j) == tb[i][1]) {
        ans[i][j] = 1;
        break;
      }
      if (cal(0, 0, j) == tb[i][0] && cal(1, 0, j) == tb[i][1]) {
        ans[i][j] = 0;
        break;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  char o[5];
  for (int i = 0; i < n; i++) {
    scanf("%s%d", o, &r[i]);
    op[i] = tran(o);
  }
  get_tb();
  get_ans();
  int val[3] = {0, 0, 0};
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 3; j++) {
      val[j] |= (ans[i][j] << i);
    }
  }
  printf("3\n| %d\n& %d\n^ %d\n", val[0], val[1], val[2]);
  return 0;
}
