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;
inline long long read() {
  long long f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
long long lowbit(long long x) { return x & -x; }
vector<pair<long long, pair<long long, long long> > > vec;
void calc(long long &x) {
  long long cc = 0, s = x;
  while (s) {
    cc++;
    s /= 2ll;
  }
  s = x;
  for (long long i = 1; i < cc; i++) {
    vec.push_back(make_pair(1, make_pair(s, s)));
    s *= 2ll;
  }
  long long y = s;
  vec.push_back(make_pair(0, make_pair(x, y)));
  long long z = x ^ y;
  vec.push_back(make_pair(1, make_pair(y, z)));
  long long p = y + z;
  vec.push_back(make_pair(0, make_pair(p, x)));
  long long q = p ^ x;
  vec.push_back(make_pair(1, make_pair(y, y)));
  long long h = 2 * y;
  vec.push_back(make_pair(0, make_pair(q, h)));
  long long e = q ^ h;
  while (e <= z) {
    if (z & e) vec.push_back(make_pair(0, make_pair(z, e))), z ^= e;
    vec.push_back(make_pair(1, make_pair(e, e))), e *= 2ll;
  }
  x = z;
  return;
}
long long X;
signed main() {
  X = read();
  while (X != 1) calc(X);
  printf("%lu\n", vec.size());
  for (auto pp : vec) {
    if (pp.first)
      printf("%lld + %lld\n", pp.second.first, pp.second.second);
    else
      printf("%lld ^ %lld\n", pp.second.first, pp.second.second);
  }
  return 0;
}