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;
const long long int maxK = 40;
long long int basis[maxK];
struct Obj {
  long long int a, b;
  char c;
  Obj() {}
  Obj(long long int a, char c, long long int b) : a(a), c(c), b(b){};
};
vector<Obj> ans;
vector<long long int> v;
bool can(long long int x) {
  for (long long int k = maxK - 1; k >= 0; k--) {
    if (!((x >> k) & 1)) continue;
    if (!basis[k]) return true;
    x ^= basis[k];
  }
  return false;
}
void insert(long long int x) {
  long long int orix = x;
  for (long long int k = maxK - 1; k >= 0; k--) {
    if (!((x >> k) & 1)) continue;
    if (!basis[k]) {
      basis[k] = x;
      return;
    }
    x ^= basis[k];
  }
}
void insert_back(long long int x) {
  long long int orix = x;
  for (long long int k = maxK - 1; k >= 0; k--) {
    if (!((x >> k) & 1)) continue;
    if (!basis[k]) {
      for (long long int l = maxK - 1; l > k; l--) {
        if (!((orix >> l) & 1)) continue;
        ans.emplace_back(orix, '^', basis[l]);
        orix ^= basis[l];
      }
      basis[k] = x;
      return;
    }
    x ^= basis[k];
  }
}
long long int x;
signed main() {
  cin >> x;
  long long int k = x;
  ans.emplace_back(x, '^', x);
  while (!basis[0]) {
    if (can(k)) {
      v.push_back(k / x);
      insert(k);
    }
    k += x;
  }
  fill(basis, basis + maxK, 0);
  for (long long int i = x; 2 * i < (1LL << maxK); i *= 2) {
    ans.emplace_back(i, '+', i);
  }
  for (long long int k : v) {
    long long int cnt = 0;
    for (long long int i = 0; i < maxK; i++) {
      if ((k >> i) & 1) {
        ans.emplace_back(cnt, '+', x * (1 << i));
        cnt += x * (1 << i);
      }
    }
    insert_back(k * x);
  }
  cout << ans.size() << endl;
  for (auto o : ans) {
    cout << o.a << " " << o.c << " " << o.b << endl;
  }
}