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 p[110];
long long a[1000000], b[1000000];
char s[1000000];
int op = 0;
map<long long, int> mp;
void add(long long x) {
  int ok = 0;
  for (int i = 55; i + 1; i--) {
    if (!(x >> i)) continue;
    if (!p[i]) {
      p[i] = x;
      ok = 1;
      break;
    }
    if (mp[x ^ p[i]] == 0) {
      ++op;
      a[op] = x, b[op] = p[i], s[op] = '^';
      mp[x ^ p[i]] = 1;
    }
    x ^= p[i];
  }
}
bool find_(long long x) {
  for (int i = 55; i + 1; i--) {
    if (!(x >> i)) continue;
    if (!p[i]) {
      return 1;
    }
    x ^= p[i];
  }
  return 0;
}
long long f[10100];
int main() {
  long long n;
  cin >> n;
  op++;
  a[op] = n, b[op] = n, s[op] = '^';
  long long m = n;
  mp[n] = 1, mp[0] = 1;
  add(n);
  f[0] = n, f[1] = n;
  int pp = 2;
  while (find_(1)) {
    n = f[pp - 2], m = f[pp - 1];
    if (mp[n + m] == 0) {
      op++;
      a[op] = n, b[op] = m, s[op] = '+';
      mp[m + n] = 1;
    }
    m += n;
    f[++pp] = m;
    add(m);
  }
  cout << op << endl;
  for (int i = 1; i <= op; i++) {
    printf("%lld %c %lld\n", a[i], s[i], b[i]);
  }
}