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 x, y;
  char t;
  Op(long long x, long long y, char t) : x(x), y(y), t(t) {}
};
vector<Op> ans;
long long mul(long long x, long long k) {
  long long res = 0;
  while (k != 0) {
    if (k & 1) {
      ans.push_back({res, x, '+'});
      res += x;
    }
    ans.push_back({x, x, '+'});
    x += x;
    k >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long x1, y1;
  long long g = gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - (a / b) * y1;
  return g;
}
void solve() {
  long long x;
  cin >> x;
  ans.push_back({x, x, 2});
  long long k = 1;
  while (k * 2 <= x) k *= 2;
  long long y = mul(x, k);
  ans.push_back({y, x, '^'});
  y ^= x;
  long long k1, k2;
  long long g = gcd(x, y, k1, k2);
  long long x1 = mul(x, abs(k1));
  long long y1 = mul(y, abs(k2));
  if ((x1 ^ y1) != 1) {
    ans.push_back({x1, x, '+'});
    x1 += x;
    ans.push_back({y1, x, '+'});
    y1 += x;
  }
  ans.push_back({x1, y1, '^'});
  cout << ans.size() << "\n";
  for (Op &op : ans) {
    cout << op.x << (op.t == '+' ? " + " : " ^ ") << op.y << "\n";
  }
}
int main() {
  int tests = 1;
  while (tests--) {
    solve();
  }
  return 0;
}