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 node {
  long long x, y, op;
  node(long long X, long long Y, long long t) {
    x = X;
    y = Y;
    op = t;
  }
};
vector<node> c;
long long calc(long long x) {
  long long y = x, ty = (x >> 1);
  while (ty) {
    c.push_back(node(y, y, 0));
    ty >>= 1, y <<= 1;
  }
  long long z = x ^ y;
  c.push_back(node(y, x, 1));
  long long rc = z + y;
  c.push_back(node(z, y, 0));
  long long we = y + y;
  c.push_back(node(y, y, 0));
  long long p = we ^ rc;
  c.push_back(node(we, rc, 1));
  long long q = p ^ x;
  c.push_back(node(p, x, 1));
  while (y != (y & -y)) {
    if (y & q) {
      c.push_back(node(y, q, 1));
      y ^= q;
    }
    c.push_back(node(q, q, 0));
    q <<= 1;
  }
  c.push_back(node(y, x, 1));
  return x ^ y;
}
signed main() {
  long long n;
  scanf("%lld", &n);
  while (n != 1) n = calc(n);
  printf("%lld\n", (long long)c.size());
  for (long long i = 0; i < (long long)c.size(); i++) {
    if (c[i].op == 0)
      printf("%lld + %lld \n", c[i].x, c[i].y);
    else
      printf("%lld ^ %lld\n", c[i].x, c[i].y);
  }
  return 0;
}