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;
const int N = 100;
const long long INF = 1e9 + 100;
void prBin(long long x) {
  vector<long long> v;
  while (x > 0) {
    v.push_back(x % 2);
    x /= 2;
  }
  int n = v.size();
  for (int i = n - 1; i >= 0; --i) {
    cout << v[i];
  }
  cout << "\n";
}
long long M = 1e4;
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int que;
void fakeGetMult(long long x, long long m) {
  if (m == 1) return;
  ++que;
  if (m % 2 == 0) {
    fakeGetMult(x, m / 2);
  } else {
    fakeGetMult(x, m - 1);
  }
}
void getMult(long long x, long long m) {
  if (m == 1) return;
  if (m % 2 == 0) {
    getMult(x, m / 2);
    cout << (x * m) / 2 << " + " << (x * m) / 2 << "\n";
  } else {
    getMult(x, m - 1);
    cout << x * (m - 1) << " + " << x << "\n";
  }
}
int main() {
  long long x;
  cin >> x;
  ++que;
  long long y = 1;
  long long i;
  for (i = 1; i <= 1e6; ++i) {
    y = (i * x) ^ x;
    if (gcd(y, x) == 1) break;
  }
  fakeGetMult(x, i);
  ++que;
  long long a = 1;
  while (a < x * y) {
    a *= 2;
  }
  long long b = a + 1;
  long long t = 1;
  long long s;
  for (s = 1; s <= x; ++s) {
    if ((a - y * s) % x == 0) {
      t = (a - y * s) / x;
      break;
    }
  }
  fakeGetMult(x, t);
  fakeGetMult(y, s);
  ++que;
  long long T = 1;
  long long second;
  for (second = 1; second <= x; ++second) {
    if ((b - y * second) % x == 0) {
      T = (b - y * second) / x;
      break;
    }
  }
  fakeGetMult(x, T);
  fakeGetMult(y, second);
  ++que;
  ++que;
  cout << que << "\n";
  cout << x << " ^ " << x << "\n";
  getMult(x, i);
  cout << i * x << " ^ " << x << "\n";
  getMult(x, t);
  getMult(y, s);
  getMult(x, T);
  getMult(y, second);
  cout << x * t << " + " << y * s << "\n";
  cout << x * T << " + " << y * second << "\n";
  cout << a << " ^ " << b << "\n";
}