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;
template <class T>
bool ckmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
void __print(int x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto z : x) cerr << (f++ ? "," : ""), __print(z);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const char nl = '\n';
string ksor = " ^ ";
string suma = " + ";
vector<string> ans;
long long ope(long long a, long long b, bool oper = 0) {
  string s = to_string(a);
  string t = to_string(b);
  string ans2;
  long long num;
  if (oper) {
    ans2 = s + ksor + t;
    num = a ^ b;
  } else {
    ans2 = s + suma + t;
    num = a + b;
  }
  ans.push_back(ans2);
  return num;
}
void mul(long long a, long long b) {
  long long tren = 0;
  while (b) {
    if (b & 1) {
      if (tren != 0) ope(tren, a);
      tren += a;
    }
    ope(a, a);
    a *= 2;
    b /= 2;
  }
}
void euclid(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  long long x1, y1;
  euclid(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
}
void solve() {
  long long x;
  cin >> x;
  long long taj = -1;
  for (int i = (1 << 30); i; i /= 2) {
    if (x & i) {
      taj = i;
      break;
    }
  }
  for (int i = 1; i < taj; i *= 2) {
    ope(x * 1LL * i, x * 1LL * i);
  }
  long long y = ope(x * taj, x, 1);
  ;
  long long a, b;
  euclid(x, y, a, b);
  while (a < 0) {
    a += y;
    b -= x;
  }
  if (b < 0) b *= -1;
  if (b % 2) {
    a += y;
    b += x;
  };
  mul(x, a);
  mul(y, b);
  ope(x * a, y * b, 1);
  cout << (int)(ans).size() << nl;
  for (auto &z : ans) {
    cout << z << nl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int testCases = 1;
  while (testCases--) solve();
}