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<string> ans;
inline string tostr(long long x) {
  char str[60];
  sprintf(str, "%lld", x);
  return (string)str;
}
namespace LB {
long long b[60];
inline bool check(long long x) {
  for (int i = 59; ~i; i--)
    if (x >> i & 1) x ^= b[i];
  return !x;
}
inline void Add(long long x) {
  if (x >= (1ll << 60)) return;
  for (int i = 59; ~i; i--)
    if (x >> i & 1) {
      if (!b[i]) {
        b[i] = x;
        break;
      }
      ans.push_back(tostr(x) + " ^ " + tostr(b[i]));
      x ^= b[i];
    }
}
}  // namespace LB
int main() {
  int x;
  scanf("%d", &x);
  LB::Add(x);
  vector<long long> vec;
  vec.push_back(x);
  while (!LB::check(1)) {
    vector<long long> nv;
    for (long long a : vec) {
      for (long long b : vec) {
        if (!LB::check(a + b)) {
          ans.push_back(tostr(a) + " + " + tostr(b));
          LB::Add(a + b);
          nv.push_back(a + b);
        }
      }
    }
    for (auto a : nv) vec.push_back(a);
  }
  printf("%d\n", (int)ans.size());
  for (auto &s : ans) puts(s.c_str());
  return 0;
}