# Min Cost To Add New Roads

There are `N`

cities numbered from `1`

to `N`

.

You are given connections, where each `connections[i] = [city1, city2, cost]`

represents the cost to connect `city1`

and `city2together`

.
(A connection is `bidirectional`

: connecting `city1`

and `city2`

is the same as connecting `city2`

and `city1`

.)

Return the minimum cost so that for every pair of cities,
there exists a path of connections (possibly of length `1`

) that connects those two cities together.
The cost is the sum of the connection costs used. If the task is impossible, return `-1`

.

**NOTE: The answer may be 0, i.e. removing the entire string.**

### Examples

#### Example 1:

##### Input: `N = 3`

, `connections = [[1,2,5],[1,3,6],[2,3,1]]`

##### Output: `6`

##### Explanation: Choosing any `2`

edges will connect all cities so we choose the `minimum 2`

.

#### Example 2:

##### Input: `N = 4`

, `connections = [[1,2,3],[3,4,4]]`

##### Output: `-1`

##### Explanation:

There is no way to connect all cities even if all edges are used.

#### Constrains:

`1 <= N <= 10000`

`1 <= connections.length <= 10000`

`1 <= connections[i][0], connections[i][1] <= N`

`0 <= connections[i][2] <= 10^5`

`connections[i][0] != connections[i][1]`

### Solution:

Try to connect cities with minimum cost, then find a small cost edge first, if two cities connected by the edge do not have the same ancestor, then union them.

When the number of unions equal to `1`

, all cities is connected.

Time Complexity: `O(mlogm + mlogN)`

. sort takes `O(mlogm)`

. find takes `O(logN)`

. With path compression and unino by weight, amatorize O(1).

Space: `O(N)`

.