DFS spanning tree

Consider a Depth-First-Search(DFS) spanning tree T of a undirected connected graph G, we define a T-Simple Circle as a path v₁, v₂, ..., v_k (v₁ = v_k) in G that contains at most one edge which not belongs to the DFS spanning tree T.
Given a graph G, we process DFS on G starting from vertex 1 and get a DFS spanning tree T, then you should choose some edges of G so that all T-Simple Circles contains at least one edge that you choose.
Please minimize the number of edges you choose.

There are at most 100 test cases.
For each case, the first line contains two integers n and m denoting the number of vertices and edges. The vertexes are numbered from 1 to n.
The following m lines describe the graph. Each line contains two integers x_i and y_i, denoting an edge between vertex x_i and y_i(x_i ≠ y_i).
Note that the first n-1 edges of input construct a DFS spanning tree T which is generated by DFS from vertex 1.
Input ends with n = 0 and m = 0
(1 <= n <= 2000, 1 <= m <= 20000, 1 <= x_i, y_i <= n)

For each case, output the number of minimal edges that you choose.

4 5
1 2
4 2
3 2
1 3
4 1
10 14
1 2
2 9
1 3
3 4
4 10
3 5
5 7
5 8
3 6
9 1
1 10
1 6
7 3
8 3
0 0

1
3

Here is the graph G in the first test case:

The solid lines denote the edge which belongs to the DFS spanning tree, and the dotted lines denote the others.
For first case, we can choose one edge (1, 2)
For second case, we can choose three edges (1, 2), (1, 3), (3, 5)