#include<bits/stdc++.h>
#define LL unsigned long long
using namespace std;
const int N=1e6+9;
const int MOD=1e9+7;
LL num[N],n,m,v,qj[N];
bool vis[N];
struct node{
	LL c,d;
};
node sj[N];
bool cmp(node x,node y){
	return x.c<y.c;
}
bool check(){
	for(int i=1;i<m;i++){
		if(sj[i].c==sj[i+1].c&&sj[i].d!=sj[i+1].d)return 1;
	}
	return 0;
}
LL ksm(LL base,LL t){
	LL s=1;
	while(t){
		if(t%2)s*=base;
		t/=2;
		base*=base;
		base%=MOD;
		s%=MOD;
	}
	return s;
}
LL find(LL x){
	if(x==0)return 1;
//	if(x==1)return qj[x]=(v*(v-1)+1)%MOD;//%MOD
//	if(qj[x])return qj[x];
//	else return qj[x]=(ksm(v*v,x-1)*v*(v-1)+v*find(x-1))%MOD;//%MOD%MOD%MOD
	LL s=(ksm(v,x*2))-(ksm(v,x))+(ksm(v,x-1));
	return s%MOD;
}
void solve(){
	LL ans=1;
	ans*=ksm(v,2*(sj[1].c-1));
	ans%=MOD;
	for(int i=2;i<=m;i++){
		ans*=find(sj[i].c-sj[i-1].c);
		ans%=MOD;
	}
	ans*=ksm(v,2*(n-sj[m].c));
	ans%=MOD;
	printf("%lld\n",ans);
}
int main(){
//	freopen("assign.in","r",stdin);
//	freopen("assign.out","w",stdout);
	LL T;
	scanf("%lld",&T);
	while(T--){
		scanf("%lld%lld%lld",&n,&m,&v);
		for(int i=1;i<=m;i++){
			scanf("%lld%lld",&sj[i].c,&sj[i].d);
		}
		sort(sj+1,sj+m+1,cmp);
		if(check()){
			printf("0\n");
			continue ;
		}
		//if(m<=1)solve1();else
		solve();
	}
	return 0;
}
//11:28
/*
1
557584238 1 464289274
309302662 462600170
*/ 
//12:03
//12:35

这是lz的赛时代码60pts

但如果将39行

LL s=(ksm(v,x*2))-(ksm(v,x))+(ksm(v,x-1));

改为

LL s=(ksm(v,x*2))+MOD-(ksm(v,x))+(ksm(v,x-1));

即可AC

蒟蒻不理解这样为什么就对了