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;
char buf[1 << 21], *p1 = buf, *p2 = buf;
template <class T>
void read(T &x) {
  x = 0;
  int c = getchar();
  int flag = 0;
  while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (flag) x = -x;
}
template <class T>
T _max(T a, T b) {
  return b < a ? a : b;
}
template <class T>
T _min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
bool checkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
bool checkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
long long x, y;
int h;
struct node {
  long long a, b;
  int t;
  node(long long a = 0, long long b = 0, int c = 0) : a(a), b(b), t(c) {}
};
vector<node> ans;
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= (a / b) * x;
  }
}
void init() {
  read(x);
  for (int i = 0; i <= 20; ++i)
    if (x >> i & 1) h = i;
  for (int i = 0; i < h; ++i) {
    ans.push_back(node(x << i, x << i, 0));
  }
  ans.push_back(node(x, x << h, 1));
  y = x ^ (x << h);
}
void getnum(long long x, long long k) {
  int h = 0;
  for (int i = 0; i <= 50; ++i)
    if (k >> i & 1) h = i;
  for (int i = 0; i < h; ++i) {
    ans.push_back(node(x << i, x << i, 0));
  }
  long long sum = (x << h);
  for (int i = h - 1; i >= 0; --i)
    if (k >> i & 1) {
      ans.push_back(node(sum, x << i, 0));
      sum += (x << i);
    }
}
void solve() {
  long long a, b;
  exgcd(x, y, a, b);
  while (a < 0) a += y, b -= x;
  if (!(a & 1)) a += y, b -= x;
  getnum(x, a);
  getnum(y, -b);
  ans.push_back(node(x * a, y * (-b), 1));
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); ++i) {
    printf("%lld %c %lld\n", ans[i].a, ans[i].t == 0 ? '+' : '^', ans[i].b);
  }
}
int main() {
  init();
  solve();
  return 0;
}