Problem: Niwel is a little golden bear. As everyone knows, bears live in forests, but Niwel got tired of seeing all the trees so he decided to move to the city.

In the city, Niwel took on a job managing bears to deliver goods. The city that he lives in can be represented as a directed graph with n nodes and m edges. Each edge has a weight capacity. A delivery consists of a bear carrying weights with their bear hands on a simple path from node 1 to node n. The total weight that travels across a particular edge must not exceed the weight capacity of that edge.

Niwel has exactly x bears. In the interest of fairness, no bear can rest, and the weight that each bear carries must be exactly the same. However, each bear may take different paths if they like.

Niwel would like to determine, what is the maximum amount of weight he can deliver (it's the sum of weights carried by bears). Find the maximum weight.

Input

The first line contains three integers n, m and x (2 ≤ n ≤ 50, 1 ≤ m ≤ 500, 1 ≤ x ≤ 100 000) — the number of nodes, the number of directed edges and the number of bears, respectively.

Each of the following m lines contains three integers ai, bi and ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1 000 000). This represents a directed edge from node ai to bi with weight capacity ci. There are no self loops and no multiple edges from one city to the other city. More formally, for each i and j that i ≠ j it's guaranteed that ai ≠ aj or bi ≠ bj. It is also guaranteed that there is at least one path from node 1 to node n.

Output

Print one real value on a single line — the maximum amount of weight Niwel can deliver if he uses exactly x bears. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

4 4 3
1 2 2
2 4 1
1 3 1
3 4 2


Output

1.5000000000


Input

5 11 23
1 2 3
2 3 4
3 4 5
4 5 6
1 3 4
2 4 5
3 5 6
1 4 2
2 5 3
1 5 2
3 2 30


Output

10.2222222222

Note

In the first sample, Niwel has three bears. Two bears can choose the path <image>, while one bear can choose the path <image>. Even though the bear that goes on the path <image> can carry one unit of weight, in the interest of fairness, he is restricted to carry 0.5 units of weight. Thus, the total weight is 1.5 units overall. Note that even though Niwel can deliver more weight with just 2 bears, he must use exactly 3 bears on this day.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr int64_t V = 805;
constexpr int64_t E = 20010;
constexpr int64_t N = V;
constexpr int64_t INF = INT64_MAX;
constexpr int64_t MULTIPLIER = 1e12;
int64_t start[V];
int64_t succ[E], cap[E], to[E];
int64_t edge_counter;
void add_edge(int64_t u, int64_t v, int64_t c) {
  cap[edge_counter] = c, to[edge_counter] = v;
  succ[edge_counter] = start[u];
  start[u] = edge_counter;
  ++edge_counter;
}
int64_t inv(int64_t e) { return e ^ 1; }
int64_t n, s, t;
int64_t lvl[N];
int64_t nextchld[N];
bool bfs() {
  for (int64_t i = 0; i < n; i++) lvl[i] = -1;
  queue<int64_t> q;
  q.push(s);
  lvl[s] = 0;
  while (!q.empty()) {
    int64_t u = q.front();
    q.pop();
    nextchld[u] = start[u];
    for (int64_t e = start[u]; ~e; e = succ[e]) {
      if (cap[e] > 0) {
        int64_t nxt = to[e];
        if (lvl[nxt] != -1) continue;
        lvl[nxt] = lvl[u] + 1;
        q.push(nxt);
      }
    }
  }
  return lvl[t] != -1;
}
int64_t aug(int64_t u, int64_t cflow) {
  if (u == t) return cflow;
  for (int64_t& i = nextchld[u]; i >= 0; i = succ[i]) {
    if (cap[i] > 0) {
      int64_t nxt = to[i];
      if (lvl[nxt] != lvl[u] + 1) continue;
      int64_t rf = aug(nxt, min(cflow, cap[i]));
      if (rf > 0) {
        cap[i] -= rf;
        cap[i ^ 1] += rf;
        return rf;
      }
    }
  }
  lvl[u] = -1;
  return 0;
}
int64_t mf() {
  int64_t tot = 0;
  while (bfs())
    for (int64_t x = aug(s, INF); x; x = aug(s, INF)) tot += x;
  return tot;
}
int64_t seen[N];
void check_reach(int64_t u) {
  if (seen[u]) return;
  seen[u] = true;
  for (int64_t e = start[u]; ~e; e = succ[e]) {
    if (cap[e] > 0) check_reach(to[e]);
  }
}
struct edge {
  int64_t u, v, w;
  edge(int64_t _u, int64_t _v, int64_t _w) : u{_u}, v{_v}, w{_w} {}
};
vector<edge> edges;
void cleanGraph() {
  fill(start, start + V, -1);
  fill(succ, succ + E, 0);
  fill(cap, cap + E, 0);
  fill(to, to + E, 0);
  edge_counter = 0;
  fill(seen, seen + N, 0);
  fill(lvl, lvl + N, 0);
  fill(nextchld, nextchld + N, 0);
}
bool canDo(int64_t weightPerBear, int64_t X) {
  cleanGraph();
  for (const edge& e : edges) {
    add_edge(e.u, e.v, e.w / weightPerBear);
    add_edge(e.v, e.u, 0);
  }
  return mf() >= X;
}
signed main() {
  int64_t N, M, X;
  cin >> N >> M >> X;
  for (int64_t i = 0; i < M; ++i) {
    int64_t src, dst, weight;
    cin >> src >> dst >> weight;
    edges.push_back(edge{src, dst, weight * MULTIPLIER});
  }
  n = N + 1;
  s = 1;
  t = N;
  int64_t lo = 0;
  int64_t hi = INT64_MAX;
  double ans;
  while (lo <= hi) {
    int64_t mid = (lo + hi) / 2;
    if (canDo(mid, X)) {
      ans = ((double)mid * (double)X) / MULTIPLIER;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  printf("%.10lf\n", ans);
  return 0;
}