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;
vector<long long> x, y;
vector<string> z;
void gcd(long long a, long long b, long long &s, long long &t) {
  long long r0 = a, r1 = b;
  long long s0 = 1, s1 = 0;
  long long t0 = 0, t1 = 1;
  while (r1 > 0) {
    long long q = r0 / r1, p;
    p = r0;
    r0 = r1;
    r1 = p - q * r1;
    p = s0;
    s0 = s1;
    s1 = p - q * s1;
    p = t0;
    t0 = t1;
    t1 = p - q * t1;
  }
  s = s0;
  t = t0;
}
void get(long long a, long long b) {
  long long p = 0;
  while (b) {
    if (b & 1) {
      x.push_back(a);
      y.push_back(p);
      z.push_back(" + ");
      p += a;
    }
    x.push_back(a);
    y.push_back(a);
    z.push_back(" + ");
    b >>= 1;
    a <<= 1;
  }
}
void out() {
  cout << x.size() << endl;
  for (int i = 0; i < x.size(); i++) {
    cout << x[i] << z[i] << y[i] << endl;
  }
  exit(0);
}
void solve() {
  long long n, l = 30;
  cin >> n;
  while (!(n >> l) & 1) {
    l--;
  }
  x.push_back(n);
  y.push_back(n);
  z.push_back(" ^ ");
  long long m = n;
  for (int i = 0; i < l; i++) {
    x.push_back(m);
    y.push_back(m);
    z.push_back(" + ");
    m <<= 1;
  }
  x.push_back(m);
  y.push_back(n);
  z.push_back(" ^ ");
  m = (m ^ n);
  long long s, t;
  gcd(m, n, s, t);
  s = abs(s);
  t = abs(t);
  if (((m * s) ^ (n * t)) != 1) {
    s += n;
    t += m;
  }
  get(m, s);
  get(n, t);
  x.push_back(m * s);
  y.push_back(n * t);
  z.push_back(" ^ ");
  out();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}