Duathlon

The street network of Byteburg has $n$ intersections and $m$ two-way street segments, and each segment joins two intersections. Byteburg was chosen to host the upcoming duathlon championship. The competition has two legs: a running leg, followed by a cycling leg.

The route is built like this. First pick three distinct intersections $s$, $c$ and $f$ as the start, change and finish stations. Then build a route that starts at $s$, goes through $c$ and ends at $f$. For safety, the route visits each intersection at most once.

Before planning the route, the mayor wants to count the triples $(s, c, f)$ for which such a route exists. Compute that number.

The first line contains the number of intersections $n$ and the number of streets $m$ ($1 \le n \le 10^5$, $1 \le m \le 2 \times 10^5$). Each of the next $m$ lines describes one street with the numbers $v_i$ and $u_i$ of the two intersections it joins ($1 \le v_i, u_i \le n$, $v_i \ne u_i$). At most one street joins a given pair of intersections.

Print the number of triples $(s, c, f)$ for which the route can be built.

In the first sample the 8 triples $(s, c, f)$ are $(1, 2, 3)$, $(1, 2, 4)$, $(1, 3, 4)$, $(2, 3, 4)$, $(3, 2, 1)$, $(4, 2, 1)$, $(4, 3, 1)$, $(4, 3, 2)$.

In the second sample the 14 triples $(s, c, f)$ are $(1, 2, 3)$, $(1, 2, 4)$, $(1, 3, 4)$, $(1, 4, 3)$, $(2, 3, 4)$, $(2, 4, 3)$, $(3, 2, 1)$, $(3, 2, 4)$, $(3, 4, 1)$, $(3, 4, 2)$, $(4, 2, 1)$, $(4, 2, 3)$, $(4, 3, 1)$, $(4, 3, 2)$.