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 x;
vector<pair<long long, pair<long long, long long>>> ans;
long long val(long long xx) { return __builtin_popcount(xx); }
long long gcdExtended(long long first, long long second, long long *x,
                      long long *y);
long long modInverse(long long first, long long m) {
  long long x, y;
  long long g = gcdExtended(first, m, &x, &y);
  if (g != 1)
    cout << "Inverse doesn't exist";
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
long long gcdExtended(long long first, long long second, long long *x,
                      long long *y) {
  if (first == 0) {
    *x = 0, *y = 1;
    return second;
  }
  long long x1, y1;
  long long gcd = gcdExtended(second % first, first, &x1, &y1);
  *x = y1 - (second / first) * x1;
  *y = x1;
  return gcd;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> x;
  for (long long jj = 0; jj < 40; jj++) {
    if (x * (1LL << jj) > (long long)1e18) {
      break;
    }
    ans.push_back({0, {x * (1LL << jj), x * (1LL << jj)}});
  }
  long long ind = -1;
  for (long long jj = 19; jj >= 0; jj--) {
    if (x & (1LL << jj)) {
      ind = jj;
      break;
    }
  }
  ans.push_back({1, {x * (1LL << ind), x}});
  long long y = x ^ (x * (1LL << ind));
  for (long long jj = 0; jj < 40; jj++) {
    if (y * (1LL << jj) > (long long)1e18) {
      break;
    }
    ans.push_back({0, {y * (1LL << jj), y * (1LL << jj)}});
  }
  long long kk = (modInverse(x, y) * x) % (x * y);
  if (kk % 2 == 0) {
    kk += x * y;
  }
  long long zz = kk / x;
  long long pp = (kk - 1) / y;
  ans.push_back({1, {x, x}});
  long long cur = 0;
  for (long long j = 0; j < 40; j++) {
    if ((1LL << j) & zz) {
      ans.push_back({0, {cur, x * (1LL << j)}});
      cur += x * (1LL << j);
    }
  }
  long long cur2 = 0;
  for (long long j = 0; j < 40; j++) {
    if ((1LL << j) & pp) {
      ans.push_back({0, {cur2, y * (1LL << j)}});
      cur2 += y * (1LL << j);
    }
  }
  ans.push_back({1, {cur, cur2}});
  cout << ans.size() << endl;
  for (auto j : ans) {
    cout << j.second.first << " ";
    if (j.first == 0) {
      cout << "+ ";
    } else {
      cout << "^ ";
    }
    cout << j.second.second << endl;
  }
  return 0;
}