#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pll pair<ll, ll>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
// int mod = 998244353;
int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<class T> ostream& operator << (ostream& os, const set<T>& s) { for (auto e : s) os << e; return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { (a += b % mod) %= mod; if (a < 0) a += mod; }
template<class T> void operator += (vector<T>& v, vector<T> v2) { rep(i, sz(v2)) v.eb(v2[i]); }


void solve();

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12); // debuged
  int T;
  T = 1;
  while (T--) solve();
}

// 先にこっち
// https://atcoder.jp/contests/abc144/submissions/15138762
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > to(n), ot(n);
  rep(i, m) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    to[a].eb(b);
    ot[b].eb(a);
  }

  vector<double> en(n), pn(n); // v to n の長さの期待値
  pn[n - 1] = 1;
  for (int v = n - 2; v >= 0; --v) {
    int deg = sz(to[v]);
    for (int u : to[v]) {
      en[v] += en[u];
      pn[v] += pn[u];
    }
    en[v] /= deg; en[v] += 1;
    pn[v] /= deg;
  }
  // cerr << pn << endl << en << endl;

  vector<double> e1(n), p1(n); // 1 to v の長さの期待値
  // 1 をゴールとする
  p1[0] = 1;
  for (int v = 1; v < n; ++v) {
    for (int u : ot[v]) {
      int deg = sz(to[u]); // v to u, 元々は u to v
      e1[v] += (e1[u] + 1) / deg;
      p1[v] += p1[u] / deg;
    }
  }
  // cerr << p1 << endl << e1 << endl;

  double ans = en[0];
  for (int v = n - 2; v >= 0; --v) { // v始点の辺
    // まず, v始点の辺すべて 1/deg to 1/(deg-1) にする
    int deg = sz(to[v]);
    if (deg == 1) continue;
    /*
    3 3
    1 2
    1 3
    2 3
    のケースでは deg = 1 の2からの辺を切れるが, もっと前を切るべき
    */

    double tmp = 0;
    vector<double> x;
    for (int u : to[v]) {
      // 期待値_{1 to v} = ∑_{1 to v のパス}確率*長さ
      // ∑_{1 to v}∑_{u to n}確率*長さ, 確率 = p_{1 to v} * 1/deg * p_{u to n}, 長さ = l_{1 to v} + 1 + l_{u to n}
      // 1/deg ∑_{1 to v}∑_{u to n}p1 * p2 * (l1 + 1 + l2)
      // 中身 = (p1l1)p2 + p1p2 + p1(p2l2) = E1p2 + p1p2 + p1E2
      x.eb(e1[v] * pn[u] + p1[v] * pn[u] + p1[v] * en[u]);
    }
    double sum = accumulate(all(x), 0.0);
    tmp = en[0] - sum * 1.0 / deg + sum * 1.0 / (deg - 1);

    // 1つだけ 1/(deg-1) to 0
    rep(i, sz(x)) {
      chmin(ans, tmp - x[i] / (deg - 1));
    }
  }
  cout << ans << endl;
}
// https://atcoder.jp/contests/abc144/submissions/8182596


Measured memory_bytes: 6016