// DFS version
#include <iostream>
#include <vector>
#include <numeric>
using namespace std;

struct ACMTree
{
	ACMTree( const vector<int>& C )
	{
		// 1. Reconstruction of the tree structure
		//    i.e., to determine the first child for each ACM

		vector<int> first_child_of( C.size() );
		for(int a=0,p=1; a<C.size(); p+=C[a],++a)
			first_child_of[a] = p;

		// 2. Labels by DFS visitation order
		//    id1 : DFS ID of each ACM
		//    id2 : DFS ID of the last descendant of each ACM

		int number_of_acm = accumulate( C.begin(), C.end(), 1 );
		id1 = vector<int>( number_of_acm );
		id2 = vector<int>( number_of_acm );

		// 3. Let's DFS!
		//    # 300,000 recursive call may happen, so you cannot DFS by recursion
		//    # Mmm...

		int dfsID = 0;

		vector<int> stk; stk.push_back(0); // 0=root
		while( !stk.empty() )
		{
			int acm = stk.back(); stk.pop_back();
			if( acm >= 0 )
			{
				id1[acm] = dfsID++;
				stk.push_back(~acm); // trick for recording id2 (processed after all children)
				if( acm < C.size() )
					for(int i=C[acm]-1; i>=0; --i)
						stk.push_back( first_child_of[acm]+i ); // push each child
			}
			else
				id2[~acm] = dfsID; // trick
		}
	}

	bool is_ancestor_descendant( int a, int b )
	{
		// 4. Now we have the DFS visitaion order for each ACM,
		//    So it is a quite easy task to determine the anc-des relation.
		return id1[a] <= id1[b] && id2[b] <= id2[a];
	}

	vector<int> id1, id2;
};

int main()
{
	int T; cin >> T;
	for( int caseNo=1; caseNo<=T; ++caseNo )
	{
		// Input (numbers of children)
		vector<int> C;
		{
			int N; cin >> N;
			while(N--) {int c; cin>>c; C.push_back(c);}
		}

		// Construct the tree
		ACMTree tree(C);

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

		int M; cin >> M;
		while(M--)
		{
			int a, b; cin >> a >> b;
			if( tree.is_ancestor_descendant(a,b) )
				count++;
		}

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