Problem: You have a blackboard and initially only an odd number x is written on it. Your goal is to write the number 1 on the blackboard.

You may write new numbers on the blackboard with the following two operations. 

  * You may take two numbers (not necessarily distinct) already on the blackboard and write their sum on the blackboard. The two numbers you have chosen remain on the blackboard. 
  * You may take two numbers (not necessarily distinct) already on the blackboard and write their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) on the blackboard. The two numbers you have chosen remain on the blackboard. 

Perform a sequence of operations such that at the end the number 1 is on the blackboard.

Input

The single line of the input contains the odd integer x (3 ≤ x ≤ 999,999).

Output

Print on the first line the number q of operations you perform. Then q lines should follow, each describing one operation. 

  * The "sum" operation is described by the line "a + b", where a, b must be integers already present on the blackboard. 
  * The "xor" operation is described by the line "a ^ b", where a, b must be integers already present on the blackboard. 

The operation symbol (+ or ^) must be separated from a, b by a whitespace.

You can perform at most 100,000 operations (that is, q≤ 100,000) and all numbers written on the blackboard must be in the range [0, 5⋅10^{18}]. It can be proven that under such restrictions the required sequence of operations exists. You can output any suitable sequence of operations.

Examples

Input


3


Output


5
3 + 3
3 ^ 6
3 + 5
3 + 6
8 ^ 9


Input


123


Output


10
123 + 123
123 ^ 246
141 + 123
246 + 123
264 ^ 369
121 + 246
367 ^ 369
30 + 30
60 + 60
120 ^ 121

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, pair<long long, long long> > > ans;
long long add(long long a, long long b) {
  ans.push_back({1, {a, b}});
  return a + b;
}
long long xo(long long a, long long b) {
  ans.push_back({2, {a, b}});
  return a ^ b;
}
void answer() {
  printf("%d\n", (int)ans.size());
  for (auto it : ans) {
    if (it.first == 1) {
      printf("%lld + %lld\n", it.second.first, it.second.second);
    } else {
      printf("%lld ^ %lld\n", it.second.first, it.second.second);
    }
  }
  exit(0);
}
int main() {
  long long x;
  scanf("%lld", &x);
  long long tmp = x;
  for (int i = 1; (x >> i); i++) {
    tmp = add(tmp, tmp);
  }
  tmp = xo(add(add(tmp, x), xo(tmp, x)), add(xo(tmp, x), xo(tmp, x)));
  long long new_x = x;
  for (int i = 1; (1LL << i) < tmp; i++) {
    if ((new_x >> i) & 1) {
      new_x = add(x * (1LL << i), new_x);
    }
  }
  while (new_x != 1) {
    if (new_x & tmp) {
      new_x = xo(new_x, tmp);
    }
    tmp = add(tmp, tmp);
  }
  answer();
  return 0;
}