On a 2D plane, we place `n`

stones at some integer coordinate points. Each coordinate point may have at most one stone.

A stone can be removed if it shares either **the same row or the same column** as another stone that has not been removed.

Given an array `stones`

of length `n`

where `stones[i] = [x`

represents the location of the _{i}, y_{i}]`i`

stone, return ^{th}*the largest possible number of stones that can be removed*.

**Example 1:**

Input:stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]Output:5Explanation:One way to remove 5 stones is as follows: 1. Remove stone [2,2] because it shares the same row as [2,1]. 2. Remove stone [2,1] because it shares the same column as [0,1]. 3. Remove stone [1,2] because it shares the same row as [1,0]. 4. Remove stone [1,0] because it shares the same column as [0,0]. 5. Remove stone [0,1] because it shares the same row as [0,0]. Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.

**Example 2:**

Input:stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]Output:3Explanation:One way to make 3 moves is as follows: 1. Remove stone [2,2] because it shares the same row as [2,0]. 2. Remove stone [2,0] because it shares the same column as [0,0]. 3. Remove stone [0,2] because it shares the same row as [0,0]. Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.

**Example 3:**

Input:stones = [[0,0]]Output:0Explanation:[0,0] is the only stone on the plane, so you cannot remove it.

**Constraints:**

`1 <= stones.length <= 1000`

`0 <= x`

_{i}, y_{i}<= 10^{4}- No two stones are at the same coordinate point.

class Solution {
public int removeStones(int[][] stones) {
}
}