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> ss;
void xr(long long a, long long b) {
  string s = to_string(a);
  s += ' ';
  s += '^';
  s += ' ';
  s += to_string(b);
  ss.push_back(s);
}
void sm(long long a, long long b) {
  string s = to_string(a);
  s += ' ';
  s += '+';
  s += ' ';
  s += to_string(b);
  ss.push_back(s);
}
void print(long long x) {
  for (long long i = 0; i < 30; i++) cout << ((x >> i) & 1);
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long x, lst;
  cin >> x;
  lst = x;
  long long b = x;
  vector<long long> pp;
  for (long long i = 0; i < 35; i++) {
    xr(x, b);
    sm((x ^ b), b);
    sm((x ^ b) + b, x);
    b = (x ^ b) + x + b;
    pp.push_back(b);
  }
  long long y = pp[pp.size() - 3];
  sm(y, y);
  sm(2 * y, y);
  xr(3 * y, pp.back());
  xr(pp[pp.size() - 2], pp[pp.size() - 3]);
  xr((3 * y) ^ pp.back(), pp[pp.size() - 2] ^ pp[pp.size() - 3]);
  long long cur = 4;
  for (long long i = 2; i < 50; i++) {
    if ((x >> i) & 1) {
      xr(x, cur);
      x = x ^ cur;
    }
    sm(cur, cur);
    cur = cur * 2;
  }
  if (x == 3) {
    ss.push_back("3 + 3");
    ss.push_back("3 ^ 6");
    ss.push_back("3 + 5");
    ss.push_back("3 + 6");
    ss.push_back("8 ^ 9");
  }
  cout << ss.size() << '\n';
  for (auto el : ss) cout << el << '\n';
  return 0;
}