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:
n = int(input());out = []
if n>>1 & 1:out.append(str(n) + " + " + str(n)) ;out.append(str(2*n) + " ^ " + str(n)) ; n = ((2*n)^n)
m = n.bit_length() - 1;a = n;b = n
for i in range(m):out.append(str(b) + " + " + (str(b)));b *= 2
out.append(str(a) + " + " + str(b));c = a + b;out.append(str(a) + " ^ " + str(b));d = a^b;out.append(str(c) + " ^ " + str(d));e = c^d;a = n;b = e
while True:
    if a==1 or b==1:break
    if a>b:a,b = b,a
    pre_a = a;A = a.bit_length() - 1;B = b.bit_length() - 1
    for i in range(B-A):out.append(str(a) + " + " + str(a));a *= 2
    out.append(str(a) + " ^ " + str(b));b = a^b;a = pre_a
print(len(out))
for s in out:print(s)