You are given two integers, `x`

and `y`

, which represent your current location on a Cartesian grid: `(x, y)`

. You are also given an array `points`

where each `points[i] = [a`

represents that a point exists at _{i}, b_{i}]`(a`

. A point is _{i}, b_{i})**valid** if it shares the same x-coordinate or the same y-coordinate as your location.

Return *the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location*. If there are multiple, return

`-1`

.The **Manhattan distance** between two points `(x`

and _{1}, y_{1})`(x`

is _{2}, y_{2})`abs(x`

._{1} - x_{2}) + abs(y_{1} - y_{2})

**Example 1:**

Input:x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]Output:2Explanation:Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.

**Example 2:**

Input:x = 3, y = 4, points = [[3,4]]Output:0Explanation:The answer is allowed to be on the same location as your current location.

**Example 3:**

Input:x = 3, y = 4, points = [[2,3]]Output:-1Explanation:There are no valid points.

**Constraints:**

`1 <= points.length <= 10`

^{4}`points[i].length == 2`

`1 <= x, y, a`

_{i}, b_{i}<= 10^{4}

class Solution {
public int nearestValidPoint(int x, int y, int[][] points) {
}
}