Fix return type of next when default parameter is provided.

**test_next.py**:
```python
z = (x*2 for x in range(10))
reveal_type(next(z, None))
```

Before:
```shell
test_next.py:2: error: Revealed type is 'builtins.int*'
```

After:
```shell
test_next.py:2: error: Revealed type is 'Union[builtins.int*, builtins.None]'
```

Author: Roy Williams

stdlib/2/__builtin__.pyi

@@ -725,7 +725,7 @@ def min(iterable: Iterable[_T], key: Callable[[_T], Any] = ...) -> _T: ...
 @overload
 def next(i: Iterator[_T]) -> _T: ...
 @overload
-def next(i: Iterator[_T], default: _T) -> _T: ...
+def next(i: Iterator[_T], default: _VT) -> Union[_T, _VT]: ...
 def oct(i: int) -> str: ...  # TODO __index__
 @overload
 def open(file: str, mode: str = 'r', buffering: int = ...) -> BinaryIO: ...

stdlib/3/builtins.pyi

@@ -743,7 +743,7 @@ def min(iterable: Iterable[_T], key: Callable[[_T], Any] = ..., default: _T = ..
 @overload
 def next(i: Iterator[_T]) -> _T: ...
 @overload
-def next(i: Iterator[_T], default: _T) -> _T: ...
+def next(i: Iterator[_T], default: _VT) -> Union[_T, _VT]: ...
 def oct(i: int) -> str: ...  # TODO __index__
 
 if sys.version_info >= (3, 6):