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 = 7e5 + 10;
vector<int> a[maxn];
vector<int> ans[5];
char s[maxn];
int vis[maxn];
int via[maxn];
int check(int u) {
  int res = 0;
  for (int i = ans[u].size() - 1; i >= 0; i--) {
    res = res * 2 + ans[u][i];
  }
  return res;
}
void judge(int x, int id) {
  while (x) {
    if (x & 1)
      a[id].push_back(1);
    else
      a[id].push_back(0);
    x >>= 1;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int flag = 0;
  int len = 0;
  int mark = 10;
  for (int i = 1; i <= n; i++) {
    int num;
    char ch;
    cin >> ch >> num;
    judge(num, i);
    s[i] = ch;
    int m = a[i].size();
    len = max(len, m);
    if (ch == '&') {
      flag = 1;
      mark = min(mark, m);
    }
  }
  for (int i = 1; i <= n; i++) {
    while (a[i].size() < len) a[i].push_back(0);
  }
  for (int i = 0; i < len; i++) {
    int res = 0;
    for (int k = 1; k <= n; k++) {
      if (s[k] == '^')
        res ^= a[k][i];
      else if (s[k] == '|')
        res |= a[k][i];
      else
        res &= a[k][i];
    }
    int rea = 1;
    for (int k = 1; k <= n; k++) {
      if (s[k] == '^')
        rea ^= a[k][i];
      else if (s[k] == '|')
        rea |= a[k][i];
      else
        rea &= a[k][i];
    }
    if (res == 0 && rea == 0) {
      ans[1].push_back(0);
      ans[2].push_back(0);
      ans[3].push_back(0);
    }
    if (res == 1 && rea == 1) {
      ans[1].push_back(1);
      ans[2].push_back(1);
      ans[3].push_back(0);
    }
    if (res == 0 && rea == 1) {
      ans[1].push_back(1);
      ans[2].push_back(0);
      ans[3].push_back(0);
    }
    if (res == 1 && rea == 0) {
      ans[1].push_back(1);
      ans[2].push_back(0);
      ans[3].push_back(1);
    }
  }
  while (ans[1].size() < mark) ans[1].push_back(1);
  printf("3\n");
  int num1 = check(1);
  int num2 = check(2);
  int num3 = check(3);
  printf("& %d\n", num1);
  printf("| %d\n", num2);
  printf("^ %d\n", num3);
  return 0;
}
