CF1731E.Graph Cost

You are given an initially empty undirected graph with $n$ nodes, numbered from $1$ to $n$ (i. e. $n$ nodes and $0$ edges). You want to add $m$ edges to the graph, so the graph won't contain any self-loop or multiple edges.

If an edge connecting two nodes $u$ and $v$ is added, its weight must be equal to the greatest common divisor of $u$ and $v$ , i. e. $\gcd(u, v)$ .

In order to add edges to the graph, you can repeat the following process any number of times (possibly zero):

• choose an integer $k \ge 1$ ;
• add exactly $k$ edges to the graph, each having a weight equal to $k + 1$ . Adding these $k$ edges costs $k + 1$ in total.

Note that you can't create self-loops or multiple edges. Also, if you can't add $k$ edges of weight $k + 1$ , you can't choose such $k$ .For example, if you can add $5$ more edges to the graph of weight $6$ , you may add them, and it will cost $6$ for the whole pack of $5$ edges. But if you can only add $4$ edges of weight $6$ to the graph, you can't perform this operation for $k = 5$ .

Given two integers $n$ and $m$ , find the minimum total cost to form a graph of $n$ vertices and exactly $m$ edges using the operation above. If such a graph can't be constructed, output $-1$ .

Note that the final graph may consist of several connected components.

Each test contains multiple test cases. The first line contains the number of test cases $t$ ( $1 \leq t \leq 10^4$ ). Description of the test cases follows.

The first line of each test case contains two integers $n$ and $m$ ( $2 \leq n \leq 10^6$ ; $1 \leq m \leq \frac{n(n-1)}{2}$ ).

It is guaranteed that the sum of $n$ over all test cases does not exceed $10^6$ .

For each test case, print the minimum cost to build the graph, or $-1$ if you can't build such a graph.

In the first test case, we can add an edge between the vertices $2$ and $4$ with $\gcd = 2$ . This is the only possible way to add $1$ edge that will cost $2$ .

In the second test case, there is no way to add $10$ edges, so the answer is $-1$ .

In the third test case, we can add the following edges:

• $k = 1$ : edge of weight $2$ between vertices $2$ and $4$ ( $\gcd(2, 4) = 2$ ). Cost: $2$ ;
• $k = 1$ : edge of weight $2$ between vertices $4$ and $6$ ( $\gcd(4, 6) = 2$ ). Cost: $2$ ;
• $k = 2$ : edges of weight $3$ : $(3, 6)$ and $(3, 9)$ ( $\gcd(3, 6) = \gcd(3, 9) = 3$ ). Cost: $3$ .

As a result, we added $1 + 1 + 2 = 4$ edges with total cost $2 + 2 + 3 = 7$ , which is the minimal possible cost.