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>
const int Maxn = 100000;
int x;
struct Answer {
  int op;
  long long a, b;
} ans[Maxn + 5];
int len;
bool vis[60];
void work(int x) {
  if (x == 1) {
    return;
  }
  int high;
  for (int i = 19; i >= 0; i--) {
    if ((x >> i) & 1) {
      high = i;
      break;
    }
  }
  long long tmp = x;
  for (int i = 0; i < high; i++) {
    ans[++len].op = 0;
    ans[len].a = tmp;
    ans[len].b = tmp;
    tmp <<= 1;
  }
  if (!vis[high + 1]) {
    ans[++len].op = 1;
    ans[len].a = tmp;
    ans[len].b = x;
    long long now = tmp ^ x;
    ans[++len].op = 0;
    ans[len].a = now;
    ans[len].b = tmp;
    long long sum = now + tmp;
    ans[++len].op = 1;
    ans[len].a = sum;
    ans[len].b = x;
    sum ^= x;
    ans[++len].op = 0;
    ans[len].a = tmp;
    ans[len].b = tmp;
    now = (tmp << 1);
    ans[++len].op = 1;
    ans[len].a = now;
    ans[len].b = sum;
    vis[high + 1] = 1;
  }
  for (int i = high + 1; (1ll << i) <= tmp; i++) {
    if (!vis[i]) {
      ans[++len].op = 0;
      ans[len].a = (1ll << (i - 1));
      ans[len].b = (1ll << (i - 1));
      vis[i] = 1;
    }
    if ((tmp >> i) & 1) {
      ans[++len].op = 1;
      ans[len].a = tmp;
      ans[len].b = (1ll << i);
      tmp ^= (1ll << i);
    }
  }
  vis[high] = 1;
  ans[++len].op = 1;
  ans[len].a = x;
  ans[len].b = (1ll << high);
  x ^= (1ll << high);
  work(x);
}
int main() {
  scanf("%d", &x);
  work(x);
  printf("%d\n", len);
  for (int i = 1; i <= len; i++) {
    if (ans[i].op == 0) {
      printf("%lld + %lld\n", ans[i].a, ans[i].b);
    } else {
      printf("%lld ^ %lld\n", ans[i].a, ans[i].b);
    }
  }
  return 0;
}