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;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long x;
  cin >> x;
  vector<long long> a;
  vector<long long> b;
  vector<char> op;
  vector<long long> basis;
  basis.push_back(x);
  auto add = [&](long long cand) {
    long long t = cand;
    for (long long x : basis) {
      int bit = 63 - __builtin_clzll(x);
      if (t & (1LL << bit)) {
        t ^= x;
      }
    }
    if (t == 0) {
      return false;
    }
    return true;
  };
  auto ops = [&](long long cand) {
    long long t = cand;
    for (long long x : basis) {
      int bit = 63 - __builtin_clzll(x);
      if (t & (1LL << bit)) {
        a.push_back(t);
        b.push_back(x);
        op.push_back('^');
        t ^= x;
      }
    }
    basis.push_back(t);
    sort(basis.begin(), basis.end());
    reverse(basis.begin(), basis.end());
  };
  while (find(basis.begin(), basis.end(), 1) == basis.end()) {
    int sz = basis.size();
    int i = rand() % sz;
    int j = rand() % sz;
    long long cand = basis[i] + basis[j];
    if (cand < 1e14) {
      if (add(cand)) {
        a.push_back(basis[i]);
        b.push_back(basis[j]);
        op.push_back('+');
        ops(cand);
      }
    }
  }
  cout << a.size() << endl;
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << " " << op[i] << " " << b[i] << "\n";
  }
  return 0;
}