/*
Tips:
1.Remember to consider the zero situation in the covering problems
2.Remember to test the special small samples
3.Binary search : which side of the interval is closed
*/

#include<bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (ll i = a; i <= ll(b); ++i)
#define REP(a) for(ll i = 0;i < a;i++)
#define REM(a,b) (((a)%(b) == 0)?(b):((a)%(b)+(b))%(b)) 

typedef long long ll;
typedef double db;
typedef array<ll,2> pr;
constexpr ll NAX = 1e5+10;
constexpr ll M = 0;
constexpr ll PINF = 0x3FFFFFFFFFFFFFFF;
constexpr ll NINF = -0x3FFFFFFFFFFFFFFF;

ll alst[NAX],xlst[NAX],tlst[NAX];
queue<ll> que[NAX];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	ll t,n,m;
	cin >> t;
	while(t--){
		cin >> n >> m;
		FOR(i,1,n){
			cin >> alst[i];
		}
		FOR(i,1,m){
			cin >> xlst[i] >> tlst[i];
			que[tlst[i]].push(xlst[i]);
		}
		FOR(i,1,n){
			que[i].push(PINF);
		}
		ll nxt = 1,cur = 0;
		priority_queue<pr,vector<pr>,greater<pr>> rem;
		FOR(i,1,n){
			ll p = que[i].front();
			que[i].pop();
			rem.push({p,alst[i]});
		}
		xlst[m+1] = PINF+1;
		while(!rem.empty()){
			pr p = rem.top();
			rem.pop();
			if(cur+p[1] < xlst[nxt]){
				cur += p[1];
			}
			else{
				ll q = que[tlst[nxt]].front();
				que[tlst[nxt]].pop();
				if(p[0] == xlst[nxt]){
					p[1] = alst[tlst[nxt]];
					p[0] = q;
					rem.push(p);
				}
				else{
					p[1] -= xlst[nxt]-cur;
					if(p[1] > 0)rem.push(p);
					rem.push({q,alst[tlst[nxt]]});
				}
				cur = xlst[nxt];
				nxt++;
			}
		}
		cout << cur << '\n';
	}
	return 0;
}