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:
def GSB(n):
	ans = 0
	for i in range(63):
		if (1<<i)&n:
			ans = i 
	return ans
def pop_count(n):
	ans = 0
	for i in range(63):
		if (1<<i)&n:
			ans += 1
	return ans

exists = []
ops = []
def add(a,b):
	assert(a in exists and b in exists)
	ops.append(f"{a} + {b}")
	exists.append(a+b)
	assert(a+b <= 5*10**18)
	return a+b
def xor(a,b):
	assert(a in exists and b in exists)
	ops.append(f"{a} ^ {b}")
	exists.append(a^b)
	assert((a^b) <= 5*10**18)
	return a^b

n = int(input())
exists.append(n)

x = n
for i in range(GSB(n)):
	x = add(x,x)

x1 = xor(n,x)
x2 = add(x1,x)
x3 = add(x,x)
x4 = xor(x2,x3)
powerOfTwo = xor(x4,n)
cleaners = [powerOfTwo]
for i in range(40):
	cleaners.append(add(cleaners[-1],cleaners[-1]))

def clean(n):
	for p in cleaners:
		if n&p:
			n = xor(n,p)
	return n

v = n
for i in range(1,20+1):
	if (1<<i)&v:
		v = clean(xor(v,n<<i))

assert(1 in exists)

print(len(ops))
for op in ops:
	print(op)