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>
int main() {
  using namespace std;
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int64_t X;
  cin >> X;
  struct op_t {
    int64_t a, b;
    char v;
  };
  vector<op_t> ops;
  ops.reserve(1e5);
  while (X > 1) {
    int L = (63 - __builtin_clzll(X));
    assert(X & (1ll << L));
    for (int i = 0; i < L; i++) {
      ops.push_back({X << i, X << i, '+'});
    }
    int64_t a = X << L;
    ops.push_back({a, X, '^'});
    int64_t b = a ^ X;
    for (int i = 0; i <= L; i++) {
      if (b & (1ll << i)) {
        ops.push_back({a, (X << i), '+'});
        ops.push_back({b, (X << i), '+'});
        a += (X << i);
        b += (X << i);
      }
      assert(!(b & (1ll << i)));
    }
    ops.push_back({a, b, '^'});
    int64_t nx = a ^ b;
    assert(nx != 0);
    assert(__builtin_clzll(nx) >= __builtin_clzll(X));
    X = nx;
  }
  cout << ops.size() << '\n';
  for (auto op : ops) {
    cout << op.a << ' ' << op.v << ' ' << op.b << '\n';
  }
  return 0;
}