There are NN cities in the country, and MMdirectional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici​. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become00. Now she wants to go to City NN, please help her calculate the minimum distance.

Input

The first line has one integer T(1 \le T\le 5)T(1≤T≤5), then following TT cases.

For each test case, the first line has three integers N, MN,M and KK.

Then the following MM lines each line has three integers, describe a road, U_i, V_i, C_iUi​,Vi​,Ci​. There might be multiple edges between uu and vv.

It is guaranteed that N \le 100000, M \le 200000, K \le 10N≤100000,M≤200000,K≤10,
0 \le C_i \le 1e90≤Ci​≤1e9. There is at least one path between City 11 and City NN.

Output

For each test case, print the minimum distance.

樣例輸入複製

1
5 6 1
1 2 2
1 3 4
2 4 3
3 4 1
3 5 6
4 5 2

樣例輸出複製

3

最開始真的是膨脹了，覺得最短路的題都刷過了，真的是被捶的頭皮發麻，原來最短路的變化這麼的多。

這個題就是可以使kt條邊的權值為0，求出1~n的最短路。

附上程式碼：

#include<stdio.h>
#include<bits/stdc++.h>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
const int maxn = 2e5 + 7;
const long long  inf = 1e18 + 7;
struct node
{
	int v;
	int w;
	int next;
}edge[2 * maxn];
struct qnode
{
	int u;
	int k;    //次數
	long long dis;
	bool operator < (const qnode &p)const {
		return dis > p.dis;
	}
	qnode(int _u, int _k, long long _dis)
	{
		u = _u;
		k = _k;
		dis = _dis;
	}
};
int head[maxn];
int tot = 0;
long long  dis[maxn][20];
int t, n, m;
int k;
void add(int u, int v, int w)
{
	edge[tot].v = v;
	edge[tot].w = w;
	edge[tot].next = head[u];
	head[u] = tot++;
	return;
}
void init()
{
	tot = 0;
	memset(head, -1, sizeof(head));
	return;
}
long long dij()
{
	priority_queue<qnode>q;

	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j <= k; j++)
		{
			dis[i][j] = inf;
		}
	}
	dis[1][0] = 0;
	q.push(qnode(1, 0, 0));
	while (!q.empty())
	{
		int u = q.top().u;
		int tmp_k = q.top().k;
		if (u == n)
		{
			return dis[u][tmp_k];
		}
		q.pop();
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].v;
			int w = edge[i].w;
			if (dis[u][tmp_k] + w < dis[v][tmp_k])
			{
				dis[v][tmp_k] = dis[u][tmp_k] + w;
				q.push(qnode(v, tmp_k, dis[v][tmp_k]));
			}
			if (tmp_k != k)
			{
				if (dis[v][tmp_k + 1] > dis[u][tmp_k])
				{
					dis[v][tmp_k+1] = dis[u][tmp_k];
					q.push(qnode(v, tmp_k + 1, dis[v][tmp_k + 1]));
				}
			}
		}
	}
	return -1;
}
int main()
{
	//freopen("C://input.txt", "r", stdin);
	scanf("%d", &t);
	while (t--)
	{
		init();
		scanf("%d%d%d", &n, &m, &k);
		for (int i = 0; i < m; ++i)
		{
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			add(u, v, w);
		}
		printf("%d\n", dij());
	}
	return 0;
}


dij加dp思想