// triple-bfs version
//   (with input validators)
#include <iostream>
#include <vector>
#include <numeric>
#include <cassert>
using namespace std;

int main()
{
	int T; cin >> T; assert( 1<=T && T<=20 );
	for( int caseNo=1; caseNo<=T; ++caseNo )
	{
		// Input
		vector<int> C;
		int N; cin >> N; assert( 1<=N && N<=300000 );
		while(N--) {int c; cin>>c; assert( 0<=c && c<=100 ); C.push_back(c);}

		// Solve
		//  # NOTE: 20M * sizeof int * 4 = 320MB memory required.
		//  #   If doubles are used for representing ranges, 480MB.
		//  #   It is possible to cut it down to 160MB, but such a detail
		//  #   should not bother contestants. So we (judges) have to allow
		//  #   480MB memory consumption.
		int nACM = accumulate(C.begin(), C.end(), 1);
		assert( 1<=nACM && nACM<=2000000 );

		vector<int> parent(nACM, -1);
		for(int a=0,id=1; a<C.size(); ++a)
			for(int c=0; c<C[a]; ++c,++id)
				parent[id] = a;

		vector<int> subtree_size(nACM, 1);
		for(int a=nACM-1; a>=1; --a)
			subtree_size[parent[a]] += subtree_size[a];

		vector<int> range_l(nACM,0), range_r(nACM,nACM);
		for(int a=0,id=1; a<C.size(); ++a)
			for(int c=0,left=range_l[a]; c<C[a]; ++c,++id)
				range_l[id] = left,
				range_r[id] = left+=subtree_size[id];

		// Answer for each query ...
		int count = 0;

		int M; cin >> M; assert( 1<=M && M<=500000 );
		while(M--)
		{
			int a, b; cin >> a >> b; assert( a!=b );
			if( range_l[a]<=range_l[b] && range_r[b]<=range_r[a] )
				count ++;
		}

		// Output
		cout << count << endl;
	}
}