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;
inline long long read() {
  long long x = 0, f = 1, c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
long long gcd(long long x, long long y) { return y == 0 ? x : gcd(y, x % y); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long t = y;
  y = x - a / b * y;
  x = t;
}
struct node {
  long long x, y, op;
};
vector<node> ans;
inline void zao(long long x, long long y) {
  y /= x;
  y--;
  long long nw = x;
  for (long long i = 0; (1ll << i) <= y; i++) {
    if ((y >> i) & 1) {
      ans.push_back((node){nw, x, 0});
      nw += x;
    }
    ans.push_back((node){x, x, 0});
    x <<= 1;
  }
}
long long n, m, x, y, xx, yy;
signed main() {
  n = read();
  for (long long i = 1; i <= 20 && !x; i++)
    for (long long j = 1; j <= 20; j++)
      if (i != j && gcd(n, (n << i) ^ (n << j)) == 1) {
        x = i;
        y = j;
        break;
      }
  for (long long i = 1, las = n; i <= max(x, y); i++) {
    ans.push_back((node){las, las, 0});
    las <<= 1;
    if (i == x) xx = las;
    if (i == y) yy = las;
  }
  ans.push_back((node){xx, yy, 1});
  m = xx ^ yy;
  exgcd(m, n, x, y);
  x = (x % n + n) % n * m;
  zao(n, x - 1);
  zao(m, x);
  if (x & 1) {
    ans.push_back((node){x - 1, x, 1});
  } else {
    ans.push_back((node){x - 1, n, 0});
    ans.push_back((node){x, n, 0});
    ans.push_back((node){x - 1 + n, x + n, 1});
  }
  cout << ans.size() << "\n";
  for (auto v : ans) {
    if (v.op)
      cout << v.x << " ^ " << v.y << "\n";
    else
      cout << v.x << " + " << v.y << "\n";
  }
  return (0 - 0);
}