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<tuple<long long, char, long long>> sol;
void mult(long long k, long long x) {
  long long p = x, act = 0;
  while (k) {
    if (k % 2) {
      if (act != 0) sol.push_back(make_tuple(act, '+', p));
      act += p;
    }
    k /= 2;
    if (k) sol.push_back(make_tuple(p, '+', p));
    p += p;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long x;
  cin >> x;
  int k = 30;
  while ((x & (1 << k)) == 0) k--;
  long long y = x << k;
  mult(1 << k, x);
  sol.push_back(make_tuple(x, '^', y));
  y ^= x;
  long long r = y % x;
  int m = 1;
  while (m * r % x != 1) m++;
  mult(m, y);
  y *= m;
  if (y % 2 == 0) {
    sol.push_back(make_tuple(x, '+', y));
    y += x;
  }
  mult((y - 1) / x, x);
  sol.push_back(make_tuple(y, '^', y - 1));
  cout << sol.size() << "\n";
  for (auto [a, b, c] : sol) cout << a << " " << b << " " << c << "\n";
  return 0;
}