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 PYTHON3):

Solution:
from collections import deque

class Dinic():
    def __init__(self, listEdge, s, t):
        self.s = s
        self.t = t
        self.graph    = {}
        self.maxCap   = 1000000
        # dict các node lân cận
       
        # e[0]: from,  e[1]: to,  e[2]: dung luong  
        for e in listEdge:
            
            if e[0] not in self.graph:
                self.graph[e[0]] = []
             
            if e[1] not in self.graph:
                self.graph[e[1]] = []
                                   #to     #cap   #reveser edge
            self.graph[e[0]].append([e[1], e[2],  len(self.graph[e[1]])])
            self.graph[e[1]].append([e[0],    0,  len(self.graph[e[0]])-1])
            
        self.N = len(self.graph.keys())
    
    def bfs(self):
        self.dist    = {}
        self.dist[self.s] = 0
        self.curIter = {node:[] for node in self.graph}
        
        Q = deque([self.s])
        
        while(len(Q) > 0):
            cur = Q.popleft()
            
            for index,e in enumerate(self.graph[cur]):
                # Chỉ add vào các node kế tiếp nếu dung lượng cạnh > 0 và chưa được visit trước đấy 
                if e[1] > 0 and e[0] not in self.dist:
                    self.dist[e[0]] = self.dist[cur] + 1
                    # add vào danh sách node kế tiếp của node hiện tại
                    self.curIter[cur].append(index)
                    Q.append(e[0])
        
        
    def findPath(self, cur, f):
        if cur == self.t:
            return f
        
        while len(self.curIter[cur]) > 0:
            indexEdge    = self.curIter[cur][-1]
            nextNode     = self.graph[cur][indexEdge][0]
            remainCap    = self.graph[cur][indexEdge][1]
            indexPreEdge = self.graph[cur][indexEdge][2]
            
            if remainCap > 0 and self.dist[nextNode] > self.dist[cur]:
                #self.next[cur] = indexEdge
                flow = self.findPath(nextNode,  min(f, remainCap))
                
                if flow > 0:
                    self.path.append(cur)
                    self.graph[cur][indexEdge][1]            -= flow
                    self.graph[nextNode][indexPreEdge][1]    += flow
                    #if cur == self.s:
                    #    print(self.path, flow)
                    return flow 
                #else:
                    #self.path.pop()
            self.curIter[cur].pop() 
            
        return 0
    
    def maxFlow(self):
        maxflow = 0
        flow = []
        
        while(True):
            self.bfs()
            
            if self.t not in self.dist:
                break
                
            while(True):
                self.path    = []
                f = self.findPath(self.s, self.maxCap)
                #print('iter', self.curIter)
                if f == 0:
                    break
                    
                flow.append(f)    
                maxflow += f
            
        return maxflow    

    # Tìm tập node thuộc S và T
    # sau khi đã tìm được max flow
    def residualBfs(self):
        Q = deque([self.s])
        side = {self.s:'s'}
        
        while(len(Q) > 0):
            cur = Q.popleft()
            
            for index,e in enumerate(self.graph[cur]):
                if e[1] > 0 and e[0] not in side:
                    Q.append(e[0])
                    side[e[0]] = 's'
        
        S = []
        T = [] 
        for x in self.graph:
            if x in side:
                S.append(x)
            else:
                T.append(x)
        return set(S), set(T)
        
def is_ok(val, flow, X):
    num = 0
    for f in flow:
        num += f // val
        
    if num >= X:
        return True
    return False   

n, m, X = map(int, input().split())
edge    = [list(map(int, input().split())) for _ in range(m)] 

l, r = 0, 1000000

while r-l>1e-9:
    #print(l, r)
    md = (r+l) / 2
    edge_ = [[u, v, w // md] for u, v, w in edge] 
    g = Dinic(edge_, 1, n)
    maxflow = g.maxFlow()
    
    if maxflow >= X:
        l=md
    else:
        r=md
        
print(X*r)