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;
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  long long p;
  exgcd(b, a % b, x, y);
  p = x;
  x = -y;
  y = -(p + (a / b) * y);
  return;
}
vector<pair<pair<long long, long long>, long long> > ans;
void calc_quickpow(long long now, long long val) {
  long long res = 1, ret = 0;
  while (now > 0) {
    if (now & 1 && ret != 0)
      ans.push_back(make_pair(make_pair(ret * val, res * val), 0)), ret += res;
    if (now & 1 && ret == 0) ret = res;
    ans.push_back(make_pair(make_pair(res * val, res * val), 0));
    res *= 2;
    now >>= 1;
  }
}
int main() {
  long long a, b;
  cin >> a;
  calc_quickpow(pow(2, 63 - __builtin_clzll(a)), a);
  b = ((long long)pow(2, 63 - __builtin_clzll(a)) * a) ^ a;
  ans.push_back(
      make_pair(make_pair(pow(2, 63 - __builtin_clzll(a)) * a, a), 1));
  long long x, y;
  exgcd(a, b, x, y);
  long long mx = (max(0ll, max((-x + b - 1) / b, (-y + a - 1) / a)));
  long long xnow = x + b * mx, ynow = y + a * mx;
  if (ynow % 2) ynow += a, xnow += b;
  calc_quickpow(xnow, a);
  calc_quickpow(ynow, b);
  ans.push_back(make_pair(make_pair(xnow * a, ynow * b), 1));
  printf("%lld\n", (long long)ans.size());
  for (int i = 0; i < (int)ans.size(); i++) {
    printf("%lld ", ans[i].first.first);
    if (ans[i].second == 0)
      putchar('+');
    else
      putchar('^');
    putchar(' ');
    printf("%lld\n", ans[i].first.second);
  }
  return 0;
}