Modify the binary search algorithm to find two positions [...7,8...] and [...8,9...] given target is 8.

```
public class Solution {
public int[] searchRange(int[] A, int target) {
int[] range = new int[2];
if(A.length==0){return range;}
range[0] = A[0]==target?0:binarySearch(A, target, true);
range[1] = A[A.length-1]==target?A.length-1:binarySearch(A, target, false);
return range;
}
private int binarySearch(int[] A, int target, boolean downOrUp){
int begin = 0, end = A.length-1;
int middle = 0;
while(begin < end){
middle = (begin+end)/2;
// find lower bound
if(A[middle] < target && A[middle+1] == target && downOrUp)
return middle+1;
if(A[middle+1] < target && downOrUp)
begin = middle+1;
if(A[middle+1] >= target && downOrUp)
end = middle;
// find upper bound
if(A[middle] == target && A[middle+1] > target && !downOrUp)
return middle;
if(A[middle] <= target && !downOrUp)
begin = middle+1;
if(A[middle] > target && !downOrUp)
end = middle;
}
return -1;
}
}
```