Given a bit string **S** i.e **containing "0" and "1" only**. You can perform **K** number of operations **(K>0)** in an iterative way, **always** **starting** **from** **K=1**, in the following manner:-

when **K=1**, remove one leftmost bit and append it to right.

when **K=2**, remove two leftmost bit one by one and append it to right one by one.

when **K=3**, remove three left most bit one by one and append it to right one by one.

when **K=4**, remove four left most bit one by one and append it to right one by one.

so on...

You have to print the minimum value of **K** **(K > 0)** so that we get our original input string **S** after K number of operations.

**Input format :-**

A sigle string S.

**Output format :-**

Print minimum value of K.

**Constraints:-**

2<=length of string<=10^5

**Example:-**

**Input 1:-**

1011

**Output:-**

7

**Input 2:-**

1101

**Output:-**

7

**Input 3:-**

11011

**Output:-**

4

**Input 4:-**

11010

**Output:-**

4

**Input 5:-**

01010

**Output:-**

4

**Input 6:-**

101010

**Output:-**

3

**Input 7:-**

111001

**Output:-**

3

Explanations:-

In 1st input S=1011, so starting from K=1,

K=1 , S=0111 (1011 -> 0111)

K=2 , S=1101 (0111 -> 1110 -> 1101)

K=3 , S=1110 (1101 -> 1011 -> 0111 -> 1110)

K=4 , S=1110 (1110 -> 1101 -> 1011 -> 0111 -> 1110)

K=5 , S=1101 (1110 -> 1101 -> 1011 -> 0111 -> 1110 -> 1101)

K=6 , S=0111 (1101 -> 1011 -> 0111 -> 1110 -> 1101 -> 1011 -> 0111)

K=7 , S=1011 (0111 -> 1110 -> 1101 -> 1011 -> 0111 -> 1110 -> 1101 -> 1011)

so answer is K=7.