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 JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.AbstractCollection;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author whhone
 */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    OutputWriter out = new OutputWriter(outputStream);
    TaskD solver = new TaskD();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskD {
    public void solve(int testNumber, InputReader in, OutputWriter out) {
      int N = in.nextInt();
      int M = in.nextInt();
      int X = in.nextInt();

      int[] a = new int[M];
      int[] b = new int[M];
      int[] c = new int[M];

      for (int i = 0; i < M; i++) {
        a[i] = in.nextInt() - 1;
        b[i] = in.nextInt() - 1;
        c[i] = in.nextInt();
      }

      double low = 0, high = 1000000;

      while ((high - low) * X > 1e-7) {
        double m = (low + high) / 2;
        MaxFlow mf = new MaxFlow(N);
        for (int i = 0; i < M; i++) {
          mf.addEdge(a[i], b[i], (int) Math.floor(c[i] / m));
        }
        int flow = mf.solve(0, N - 1);

        if (flow >= X) {
          low = m;
        } else {
          high = m;
        }
      }

      out.println((low + high) * X / 2);
    }

  }

  static class OutputWriter extends PrintWriter {
    public OutputWriter(OutputStream stream) {
      super(stream);
    }

    public OutputWriter(Writer writer) {
      super(writer);
    }

  }

  static class MaxFlow {
    int N;
    public ArrayList<ArrayList<MaxFlowEdge>> edges;

    public MaxFlow(int N) {
      this.N = N;
      this.edges = new ArrayList<ArrayList<MaxFlowEdge>>(N);
      for (int i = 0; i < this.N; i++) {
        this.edges.add(new ArrayList<MaxFlowEdge>());
      }
    }

    public void addEdge(int from, int to, int cap) {
      MaxFlowEdge ef = new MaxFlowEdge(from, to, cap);
      MaxFlowEdge et = new MaxFlowEdge(to, from, 0);
      ef.reverse = et;
      et.reverse = ef;
      this.edges.get(from).add(ef);
      this.edges.get(to).add(et);
    }

    private int[] build(int starting) {
      int[] dist = new int[this.N];
      Arrays.fill(dist, -1);

      dist[starting] = 0;
      LinkedList<Integer> queue = new LinkedList<Integer>();

      queue.add(starting);
      while (!queue.isEmpty()) {
        int cur = queue.removeFirst();
        for (MaxFlowEdge edge : edges.get(cur)) {
          if (edge.cap > 0 && dist[edge.getTo()] == -1) {
            dist[edge.getTo()] = dist[cur] + 1;
            queue.add(edge.getTo());
          }
        }
      }

      return dist;
    }

    private int push(int u, int T, int input, int[] dist) {
      if (u == T) {
        return input;
      }
      int used = 0;
      for (MaxFlowEdge edge : edges.get(u)) {
        if (input - used <= 0) {
          break;
        }
        int v = edge.getTo();
        if (edge.cap > 0 && dist[v] == dist[u] + 1) {
          int t = push(v, T, Math.min(input - used, edge.cap), dist);
          if (t > 0) {
            edge.cap -= t;
            edge.reverse.cap += t;
            used += t;
          }
        }
      }
      if (used == 0) {
        dist[u] = -1;
      }
      return used;
    }

    public int solve(int S, int T) {
      int flow = 0;
      while (true) {
        int[] dist = build(S);
        if (dist[T] == -1) {
          break;
        }
        int tmp = push(S, T, Integer.MAX_VALUE, dist);
        flow += tmp;
      }
      return flow;
    }

    class MaxFlowEdge extends Edge {
      public int cap;
      public MaxFlowEdge reverse;

      public MaxFlowEdge(int from, int to, int cap) {
        super(from, to);
        this.cap = cap;
      }

    }

  }

  static class Edge {
    private final int from;
    private final int to;

    public Edge(int from, int to) {
      this.from = from;
      this.to = to;
    }

    public int getTo() {
      return this.to;
    }

  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;
    public String cache;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
      cache = null;
    }

    private String nextNoCache() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        tokenizer = new StringTokenizer(nextLine());
      }
      return tokenizer.nextToken();
    }

    public String nextLine() {
      try {
        return reader.readLine();
      } catch (IOException e) {
        throw new RuntimeException(e);
      }
    }

    public String next() {
      if (cache != null) {
        String next = cache;
        cache = null;
        return next;
      } else {
        return nextNoCache();
      }
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

  }
}