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;
long long x, y, X, Y;
vector<string> ans;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
string toStr(long long a) {
  string s = "";
  while (a) {
    s.push_back('0' + (a % 10));
    a /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
void addXor(long long a, long long b) {
  ans.push_back(toStr(a) + " ^ " + toStr(b));
}
void addSum(long long a, long long b) {
  ans.push_back(toStr(a) + " + " + toStr(b));
}
void handleMult(long long a, long long b) {
  if (a == b) return;
  long long mult = b / a;
  int k;
  for (k = 1; (1LL << k) <= mult; k++) {
    long long K = (1LL << (k - 1));
    long long Ka = K * a;
    addSum(Ka, Ka);
  }
  long long curr = (a << (--k));
  for (k = k - 1; k >= 0; k--)
    if (mult & (1LL << k)) {
      long long add = (a << k);
      addSum(curr, add);
      curr += add;
    }
}
int main() {
  scanf("%lld", &x);
  for (int k = 1; true; k++) {
    long long K = (1LL << k);
    long long Kx = K * x;
    y = (Kx ^ x);
    if (gcd(x, y) == 1) {
      handleMult(x, Kx);
      addXor(Kx, x);
      break;
    }
  }
  for (long long mult = 1; true; mult++) {
    Y = y * mult;
    if (Y % 2)
      X = Y - 1;
    else
      X = Y + 1;
    if (X % x == 0) {
      handleMult(x, X);
      handleMult(y, Y);
      addXor(X, Y);
      break;
    }
  }
  int cnt = ans.size();
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++) printf("%s\n", ans[i].c_str());
  return 0;
}