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 int N = 1000005;
const int inf = 1000 * 1000 * 1000;
const int mod = 1000 * 1000 * 1000 + 7;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
long long n;
struct node {
  long long x, y;
  char op;
};
vector<node> answ;
vector<int> mas, ms;
int used[N];
int main() {
  cin >> n;
  for (int i = 20; i >= 0; i--) {
    if ((n >> i) & 1) {
      mas.push_back(i);
    }
  }
  long long x = n;
  for (int i = 1; i <= mas[0]; i++) {
    answ.push_back({x, x, '+'});
    x *= 2ll;
  }
  answ.push_back({x, n, '^'});
  x ^= n;
  long long t = x % n;
  int cnt = 0;
  while (used[t] == 0) {
    used[t] = ++cnt;
    t += x;
    t %= n;
  }
  assert(used[n - 1] != 0);
  for (int i = 20; i >= 0; i--) {
    if ((used[n - 1] >> i) & 1) {
      ms.push_back(i);
    }
  }
  long long y = x;
  for (int i = 1; i <= ms[0]; i++) {
    answ.push_back({y, y, '+'});
    y *= 2ll;
  }
  long long cur = y;
  if ((int)ms.size() > 1) {
    for (int i = 1; i < (int)ms.size(); i++) {
      answ.push_back({cur, (x << ms[i]), '+'});
      cur += (x << ms[i]);
    }
  }
  long long d = (cur + 1) / n;
  mas.clear();
  for (int i = 60; i >= 0; i--) {
    if ((d >> i) & 1) {
      mas.push_back(i);
    }
  }
  x = n;
  for (int i = 1; i <= mas[0]; i++) {
    answ.push_back({x, x, '+'});
    x *= 2ll;
  }
  y = x;
  if ((int)mas.size() > 1) {
    for (int i = 1; i < (int)mas.size(); i++) {
      answ.push_back({y, (n << mas[i]), '+'});
      y += (n << mas[i]);
    }
  }
  assert(y == cur + 1);
  if (cur % 2) {
    answ.push_back({cur, n, '+'});
    answ.push_back({cur + 1, n, '+'});
    answ.push_back({cur + n, cur + n + 1, '^'});
  } else {
    answ.push_back({cur, cur + 1, '^'});
  }
  printf("%d\n", (int)answ.size());
  for (int i = 0; i < (int)answ.size(); i++) {
    cout << answ[i].x << " " << answ[i].op << " " << answ[i].y << endl;
  }
  return 0;
}