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;
struct op {
  long long a;
  long long b;
  bool is_xor;
};
const bool ADD = false;
const bool XOR = true;
void go(long long x) {
  vector<op> ret;
  while (x != 1) {
    int xlen = 0;
    long long xx = x;
    while (xx) {
      xx >>= 1;
      xlen++;
    }
    xx = x;
    for (int i = 0; i < xlen - 1; ++i) {
      ret.push_back(op{xx, xx, false});
      xx <<= 1;
    }
    long long p1 = x + xx;
    long long p2 = x ^ xx;
    long long p3 = p1 ^ p2;
    long long xx_prime = xx ^ p3;
    long long x_plus_xx_prime = x + xx_prime;
    long long y_spot = x_plus_xx_prime ^ p1;
    assert(p3 == (1LL << (xlen)) || y_spot == (1LL << (xlen)));
    ret.push_back(op{x, xx, ADD});
    ret.push_back(op{x, xx, XOR});
    ret.push_back(op{p1, p2, XOR});
    ret.push_back(op{xx, p3, XOR});
    ret.push_back(op{x, xx_prime, ADD});
    ret.push_back(op{x_plus_xx_prime, p1, XOR});
    y_spot = (1LL << (xlen));
    while (y_spot < xx) {
      if ((xx & y_spot) != 0) {
        ret.push_back(op{y_spot, xx, XOR});
        xx ^= y_spot;
      }
      ret.push_back(op{y_spot, y_spot, ADD});
      y_spot <<= 1;
    }
    assert(xx == (1LL << (xlen - 1)));
    ret.push_back(op{x, xx, XOR});
    x ^= xx;
  }
  cout << ret.size() << endl;
  for (int i = 0; i < ret.size(); ++i) {
    cout << ret[i].a << " " << (ret[i].is_xor == ADD ? "+" : "^") << " "
         << ret[i].b << endl;
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  long long x;
  cin >> x;
  go(x);
}