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;
int ans;
int m, n, maxflow;
const int length = 1000;
const int mymax = -1;
int path[2000];
int id[2000];
int thetop = 0;
int a[505][3];
class mincostMaxFlow {
 private:
  int flow[length * 2];
  int capacity[length * 2];
  int value[length * 2];
  int head[length];
  int next[length];
  bool isvisited[length * 2];
  int queue[length * 5];
  int dist[length];
  int cost[length];
  int pos;
  int e[length];
  int prev[length];
  int source;
  int target;
  int prev_vec[length];
  double smallest;

 public:
  void init(int s, int e) {
    source = s;
    target = e;
    pos = 0;
    clear();
  }
  void clear() {
    memset(head, -1, sizeof(head));
    memset(flow, 0, sizeof(flow));
    memset(capacity, 0, sizeof(capacity));
    pos = 0;
  }
  void reset(double d) {
    for (int i = 0; i < pos; i++) {
      flow[i] = 0.0;
    }
    smallest = d;
  }
  void add(int s, int t, int cap, int v) {
    e[pos] = t;
    cost[pos] = v;
    capacity[pos] = (double)cap;
    next[pos] = head[s];
    head[s] = pos++;
    e[pos] = s;
    cost[pos] = -v;
    next[pos] = head[t];
    head[t] = pos++;
  }
  bool find() {
    memset(isvisited, false, sizeof(isvisited));
    int start = 0;
    int end = 1;
    queue[start] = source;
    int l = 0;
    while (start < end) {
      int t = queue[start++];
      for (int i = head[t]; i != -1; i = next[i]) {
        if (capacity[i] - flow[i] > 0 && !isvisited[e[i]]) {
          prev[e[i]] = i;
          queue[end++] = e[i];
          isvisited[e[i]] = true;
          if (e[i] == target) return true;
        }
      }
    }
    return false;
  }
  int getmaxflow_mincost() {
    int i = 0;
    int j = 1;
    ans = 0;
    i = 0;
    ;
    int l = 0;
    while (find()) {
      int max = INT_MAX;
      int t = target;
      while (t != source) {
        int i = prev[t];
        if (max > capacity[i] - flow[i]) {
          max = capacity[i] - flow[i];
        }
        t = e[i ^ 1];
      }
      ans += max;
      t = target;
      while (t != source) {
        int i = prev[t];
        flow[i] += max;
        flow[i ^ 1] = -flow[i];
        prev_vec[t] = e[i ^ 1];
        t = e[i ^ 1];
      }
      j++;
    }
    return ans;
  }
  void getcost() {
    int s, t;
    for (int i = 0; i < pos; i++) {
      if (flow[i] > 0) {
        t = e[i] / 2;
        s = e[i ^ 1] / 2;
        path[t] = s;
      }
    }
    int cnt = 1;
    for (int i = 0; i < n; i++) {
      if (path[i] >= n) {
        id[i] = cnt++;
      } else {
        id[i] = id[path[i]];
      }
    }
  }
};
mincostMaxFlow solution;
void process() {
  int i, j, t;
  int ret = 0;
}
void input() {
  int i, j, t, k, u, v;
  int s;
  scanf("%d %d %d", &n, &m, &thetop);
  t = 0;
  solution.init(1, n);
  for (i = 0; i < m; i++) {
    scanf("%d %d %d", &a[i][0], &a[i][1], &a[i][2]);
    if (a[i][2] > t) t = a[i][2];
  }
  double l = 0;
  double h = (double)t;
  double mid;
  double diff = 0.00000000001;
  int c = 0;
  while (l + diff < h) {
    mid = (l + h) / 2;
    solution.clear();
    for (i = 0; i < m; i++) {
      k = (int)((double)a[i][2] / mid);
      if (k > 0)
        solution.add(a[i][0], a[i][1], k, 0);
      else if (k == -2147483648)
        solution.add(a[i][0], a[i][1], INT_MAX, 0);
    }
    v = solution.getmaxflow_mincost();
    if (v >= thetop) {
      l = mid;
    } else {
      h = mid;
    }
    c++;
    if (c >= 100) break;
  }
  printf("%lf\n", l * thetop);
}
int main() {
  input();
  process();
  return 0;
}