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 PYTHON3):

Solution:
from sys import stdin

output = []
def queue(stri):
    output.append(stri)

n = int(stdin.readline())

digits = len(bin(n))-2
cpy = n
for _ in range(digits-1):
    queue(f'{cpy} + {cpy}')
    cpy *= 2

m = n ^ cpy
queue(f'{n} ^ {cpy}')


p = n*m-1
while p % n != 0:
    p -= m

def foo(number, target):
    # gets target from number using additions only
    current = number
    while current * 2 <= target:
        queue(f'{current} + {current}')
        current *= 2

    total = current
    while total < target:
        if total + current <= target:
            queue(f'{total} + {current}')
            total += current
        current //= 2

foo(m, p+1)
foo(n, p)

if p % 2 == 0:
    queue(f'{p+1} ^ {p}')
else:
    queue(f'{p} + {n}')
    queue(f'{p+1} + {n}')
    queue(f'{p+n+1} ^ {p+n}')

print(len(output))
for s in output:
    print(s)