#include <iostream>
#include <stdio.h>
#include <algorithm>
#define MAXN 100010
#define int long long
#define forr(i,l,r) for(int i=(l);i<=(r);i++)
using namespace std;
void read(int &x){
	char ch = getchar();
	int w = 1; x = 0;
	while( ch < '0' || '9' < ch ){
		if( ch == '-' ) w = -1;
		ch = getchar();
	}
	while( '0' <= ch && ch <= '9' )
		x = x*10+ch-'0', ch = getchar();
	x *= w;
	return ;
}
const int mod = 1000000007;
int fpow(int a,int b){
	int c = 1;
	while(b){
		if(b&1) c = c*a%mod;
		a = a*a%mod;
		b >>= 1;
	}
	return c;
}
struct LIMIT{ int c,d; } p[MAXN];
bool operator <(LIMIT a,LIMIT b){ return a.c < b.c; }
int t,n,m,v,sum,ans;
signed main(){
//	freopen("assign3.in","r",stdin);
	freopen("assign.in","r",stdin);
	freopen("assign.out","w",stdout);
	read(t);
	while(t--){
		read(n), read(m), read(v);
		forr(i,1,m)
			read(p[i].c), read(p[i].d);
		sort(p+1,p+m+1);
		
		ans = 1;
		forr(i,1,m){
			if( p[i].c == p[i-1].c ){
				if( p[i].d != p[i-1].d )
				{ ans = 0; break; }
			}
			else{
				if( i == 1 ){
					sum = fpow(v,(p[i].c-p[i-1].c-1)*2)%mod;
					ans = ans*sum%mod;
				}
				else{
					sum = fpow(v,(p[i].c-p[i-1].c)*2)%mod;
					sum = (sum-fpow(v,p[i].c-p[i-1].c-1)*(v-1))%mod;
					ans = ans*sum%mod;
				}
			}
//			printf("[%lld] %lld\n",i,sum);
		}
		ans = ans*fpow(v,(n-p[m].c)*2)%mod;
		
		printf("%lld\n",(ans%mod+mod)%mod);
	}
	return 0;
}