using namespace std;
const int MAXN = 5005;
vector<pair<int, int>> g[MAXN];
int dfn[MAXN], low[MAXN], timer;
bool isBridge[20005];
int comp[MAXN];

void tarjan(int u, int peid) {
    dfn[u] = low[u] = ++timer;
    for (auto e : g[u]) {
        int v = e.first, eid = e.second;
        if (eid == peid) continue;          // 跳过同一条边的反向，重边不会误判
        if (!dfn[v]) {
            tarjan(v, eid);
            low[u] = min(low[u], low[v]);
            if (low[v] > dfn[u]) isBridge[eid] = true;
        } else {
            low[u] = min(low[u], dfn[v]);
        }
    }
}

int main() {
    int F, R;
    scanf("%d %d", &F, &R);
    for (int i = 0; i < R; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        g[a].push_back({b, i});
        g[b].push_back({a, i});
    }
    for (int i = 1; i <= F; i++)
        if (!dfn[i]) tarjan(i, -1);

    // 删掉所有桥，剩下的连通块即边双连通分量
    int cid = 0;
    for (int i = 1; i <= F; i++) {
        if (comp[i]) continue;
        cid++;
        queue<int> q;
        q.push(i); comp[i] = cid;
        while (!q.empty()) {
            int u = q.front(); q.pop();
            for (auto e : g[u]) {
                int v = e.first, eid = e.second;
                if (isBridge[eid] || comp[v]) continue;
                comp[v] = cid;
                q.push(v);
            }
        }
    }

    // 桥树中度数：每条桥给其两端分量各 +1
    vector<int> deg(cid + 1, 0);
    for (int u = 1; u <= F; u++)
        for (auto e : g[u])
            if (isBridge[e.second]) deg[comp[u]]++;

    int leaves = 0;
    for (int i = 1; i <= cid; i++)
        if (deg[i] == 1) leaves++;
    printf("%d\n", (leaves + 1) / 2);
    return 0;
}