Given a** directed acyclic graph**, with `n`

vertices numbered from `0`

to `n-1`

, and an array `edges`

where `edges[i] = [from`

represents a directed edge from node _{i}, to_{i}]`from`

to node _{i}`to`

._{i}

Find *the smallest set of vertices from which all nodes in the graph are reachable*. It's guaranteed that a unique solution exists.

Notice that you can return the vertices in any order.

**Example 1:**

Input:n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]Output:[0,3]Explanation:It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].

**Example 2:**

Input:n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]Output:[0,2,3]Explanation:Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.

**Constraints:**

`2 <= n <= 10^5`

`1 <= edges.length <= min(10^5, n * (n - 1) / 2)`

`edges[i].length == 2`

`0 <= from`

_{i,}to_{i}< n- All pairs
`(from`

are distinct._{i}, to_{i})