#include <bits/stdc++.h>

using namespace std;

const int MAXN=4e5+50;

int N,M1,M2;
long long K;

struct Edge
{
	int x,y,Len;
}e1[MAXN],e2[MAXN];
bool cmp(Edge a,Edge b)
{
	return a.Len<b.Len;
}

/*
Mp[i][j]:同时在1的i子树和2的j子树的点数量 
当一个点在另外一个图中跟Mp[i][j]个点联通，那么在这个图中就应该去掉这些贡献

合并：按秩合并
分裂：记录合并时候的Size
每次弹出最后一个
最后要reverse 
*/
map<int,int>Mp[MAXN];

vector<int>Vec1[MAXN],Vec2[MAXN];

int father1[MAXN],father2[MAXN];
int Last[MAXN],Root[MAXN],Size[MAXN];

bool Use[MAXN];


long long Merge(int i)
{
	int fx=father1[e1[i].x],fy=father1[e1[i].y];
	if(fx==fy)
	{
		return 0;
	}
	if(Vec1[fx].size()<Vec1[fy].size())
	{
		swap(fx,fy);
	}
	long long Now=1ll*Vec1[fx].size()*Vec1[fy].size();
	for(int j=0;j<Vec1[fy].size();j++)
	{
		int v=Vec1[fy][j];
		Now-=Mp[fx][father2[v]];
	}
	for(int j=0;j<Vec1[fy].size();j++)
	{
		int v=Vec1[fy][j];
		Vec1[fx].push_back(v);
		father1[v]=fx;
		Mp[fy][father2[v]]--;
		Mp[fx][father2[v]]++;
	}
	return Now;
}
long long Split(int i)
{
	if(Root[i]==0)
	return 0;
	int i1=Root[i],i2=Last[i];
	long long Now=1ll*(Vec2[i1].size()-Vec2[i2].size())*Vec2[i2].size(); 
	for(int j=0;j<Vec2[i2].size();j++)
	{
		int v=Vec2[i2][j];
		Mp[father1[v]][i1]--;
		Mp[father1[v]][i2]++;
		father2[v]=i2;
	}
	for(int j=0;j<Vec2[i2].size();j++)
	{
		int v=Vec2[i2][j];
		Now-=Mp[father1[v]][i1];
	}
	return Now;
}

int main()
{
	scanf("%d%d%d%lld",&N,&M1,&M2,&K); 
	for(int i=1;i<=M1;i++)
	{
		scanf("%d%d%d",&e1[i].x,&e1[i].y,&e1[i].Len); 
	}
	for(int i=1;i<=M2;i++)
	{
		scanf("%d%d%d",&e2[i].x,&e2[i].y,&e2[i].Len); 
	}
	sort(e1+1,e1+M1+1,cmp); 
	sort(e2+1,e2+M2+1,cmp);
	for(int i=1;i<=N;i++)
	{
		father2[i]=i;
		Vec2[i].push_back(i); 
	}
	for(int i=1;i<=M2;i++)
	{
		int fx=father2[e2[i].x],fy=father2[e2[i].y];
		if(fx==fy)
		{
			Root[i]=0;
			continue;
		}
		if(Vec2[fx].size()<Vec2[fy].size())
		{
			swap(fx,fy);
		}
		Root[i]=fx;
		Last[i]=fy;
		Size[i]=Vec2[fy].size();
		for(int j=0;j<Vec2[fy].size();j++)
		{
			Vec2[fx].push_back(Vec2[fy][j]);
			father2[Vec2[fy][j]]=fx;
		}
	}
	
	long long Cnt=0; 
	
	for(int i=1;i<=N;i++)
	{
		Mp[i][father2[i]]++;
		father1[i]=i;
		Vec1[i].push_back(i);
		if(Use[father2[i]]==false)
		{
			Use[father2[i]]=true;
			Cnt+=1ll*Vec2[father2[i]].size()*(Vec2[father2[i]].size()-1)/2;
		}
	}
	int r=M2;
	int ans=2e9+2;
	if(Cnt>=K)
	{
		ans=e2[M2].Len;
	}
	int Now=Split(r);
	while(r&&Cnt-Now>=K)
	{
		Cnt-=Now;
		r--;
		if(r)
		Now=Split(r);
	}
	if(Cnt>=K)
	ans=min(ans,e2[r].Len);
	for(int i=1;i<=M1;i++)
	{
		Cnt+=Merge(i);
		while(r&&Cnt-Now>=K)
		{
			Cnt-=Now;
			r--;
			//cout<<r<<endl;
			if(r)
			Now=Split(r);
		}
		if(Cnt>=K)
		{
			ans=min(ans,e1[i].Len+e2[r].Len);
		}
	}
	if(ans==2e9+2)
	{
		puts("-1");
	}
	else
	{
		printf("%d",ans);
	}
}

/*
stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* i and j are misplaced
	* variable misuse
	* do smth instead of nothing and stay organized
	* vector cause MLE or TLE
	* wrong file name
	* DON'T SLEEP IN THE EXAM
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
	* DON'T GET A SWELLED HEAD
	* YOU MUSTN'T HAVE EMOTION
->be based on Benq's code
*/