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;
long long x, x2, p, z, px1, px2, a;
vector<pair<pair<long long, char>, long long>> sol;
vector<long long> v1, v2;
int main() {
  cin >> x;
  if ((x & (1 << 1)) > 0) {
    sol.push_back({{x, '+'}, x});
    sol.push_back({{x, '^'}, x * 2});
    x = x ^ (x * 2);
  }
  for (int i = 0; i <= 20; i++) {
    if ((x & (1 << i)) > 0) {
      z = i;
    }
  }
  x2 = x;
  v1.push_back(x2);
  for (int i = 0; i <= z; i++) {
    sol.push_back({{x2, '+'}, x2});
    x2 *= 2;
    v1.push_back(x2);
  }
  p = x;
  for (int i = 1; i <= z; i++) {
    if ((p & (1 << i)) > 0) {
      sol.push_back({{p, '^'}, v1[i]});
      p = p ^ v1[i];
    }
  }
  sol.push_back({{p, '+'}, x});
  px1 = p + x;
  sol.push_back({{p, '^'}, x});
  px2 = p ^ x;
  sol.push_back({{px1, '^'}, px2});
  v2.push_back(px1 ^ px2);
  for (int i = 0; i < z; i++) {
    a = v2.back();
    sol.push_back({{a, '+'}, a});
    v2.push_back(a + a);
  }
  for (int i = 1; i <= z; i++) {
    if ((x & (1 << i)) > 0) {
      a = v2[i - 1];
      sol.push_back({{x, '^'}, a});
      x = x ^ a;
    }
  }
  cout << sol.size() << endl;
  for (auto e : sol) {
    cout << e.first.first << " " << e.first.second << " " << e.second << endl;
  }
}