#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef pair<int, int> PII;
const int N = 1000010;
int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1)
            res = res * a % p;
        a = a * a % p;
        k >>= 1;
    }
    return res;
}
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
int n, m;
int a[N];
struct node
{
    int l, r;
} b[N];
int dp[N];
int pre[N];
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        dp[i] = 0;
        pre[i] = 0;
    }
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        cin >> b[i].l >> b[i].r;
        pre[b[i].r + 1] = max(pre[b[i].r + 1], b[i].l); // 最远能从那个点转移过来  两个点之间不能有完整的区间
    }
    for (int i = 1; i <= n + 1; i++)
        pre[i] = max(pre[i], pre[i - 1]);
    // sort(b + 1, b + 1 + n, cmp);
    dp[0] = 0;
    deque<int> heap;
    heap.push_back(0);
    for (int i = 1; i <= n + 1; i++)
    {
        while (heap.size() && heap.front() < pre[i])
            heap.pop_front();
        dp[i] = dp[heap.front()] + a[i];
        while (heap.size() && dp[i] <= dp[heap.back()])
        {
            heap.pop_back();
        }
        heap.push_back(i);
    }
    cout << dp[n + 1] << endl;
}
signed main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}