You play a computer game. In this game, you lead a party of m heroes, and you have to clear a dungeon with n monsters. Each monster is characterized by its power a_i. Each hero is characterized by his power p_i and endurance s_i.

The heroes clear the dungeon day by day. In the beginning of each day, you choose a hero (exactly one) who is going to enter the dungeon this day.

When the hero enters the dungeon, he is challenged by the first monster which was not defeated during the previous days (so, if the heroes have already defeated k monsters, the hero fights with the monster k + 1). When the hero fights the monster, there are two possible outcomes:

  * if the monster's power is strictly greater than the hero's power, the hero retreats from the dungeon. The current day ends; 
  * otherwise, the monster is defeated. 



After defeating a monster, the hero either continues fighting with the next monster or leaves the dungeon. He leaves the dungeon either if he has already defeated the number of monsters equal to his endurance during this day (so, the i-th hero cannot defeat more than s_i monsters during each day), or if all monsters are defeated — otherwise, he fights with the next monster. When the hero leaves the dungeon, the current day ends.

Your goal is to defeat the last monster. What is the minimum number of days that you need to achieve your goal? Each day you have to use exactly one hero; it is possible that some heroes don't fight the monsters at all. Each hero can be used arbitrary number of times.

Input

The first line contains one integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of monsters in the dungeon.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the power of the i-th monster.

The third line contains one integer m (1 ≤ m ≤ 2 ⋅ 10^5) — the number of heroes in your party.

Then m lines follow, each describing a hero. Each line contains two integers p_i and s_i (1 ≤ p_i ≤ 10^9, 1 ≤ s_i ≤ n) — the power and the endurance of the i-th hero.

It is guaranteed that the sum of n + m over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case print one integer — the minimum number of days you have to spend to defeat all of the monsters (or -1 if it is impossible).

Example

Input


2
6
2 3 11 14 1 8
2
3 2
100 1
5
3 5 100 2 3
2
30 5
90 1


Output


5
-1
#include <bits/stdc++.h>
using namespace std;
long long modulo(long long base, long long exponent, long long modulus);
long long choose(long long n, long long k);
long long inverse(long long a, long long m);
void build();
void fileio();
long long ncr(long long n, long long r);
const int nax = 1e6 + 10;
const int LG = log2(nax) + 1;
vector<long long> fact(nax);
map<long long, long long> freq;
vector<long long> tree(4 * nax), tree2(4 * nax);
void update(long long node, long long nl, long long nr, long long ind,
            long long ele, vector<long long> &tree) {
  if (nl == nr) {
    tree[node] = ele;
    return;
  }
  long long mid = (nl + nr) / 2;
  if (nl <= ind && ind <= mid)
    update(2 * node + 1, nl, mid, ind, ele, tree);
  else
    update(2 * node + 2, mid + 1, nr, ind, ele, tree);
  tree[node] = max(tree[2 * node + 1], tree[2 * node + 2]);
}
long long querry(long long node, long long nl, long long nr, long long l,
                 long long r, vector<long long> &tree) {
  if (l > r || nl > nr || l > nr || nl > r) return 0;
  if (l <= nl && nr <= r) return tree[node];
  long long mid = (nl + nr) / 2;
  long long p1 = querry(2 * node + 1, nl, mid, l, r, tree);
  long long p2 = querry(2 * node + 2, mid + 1, nr, l, r, tree);
  return max(p1, p2);
}
void update2(long long node, long long nl, long long nr, long long ind,
             long long ele, vector<long long> &tree) {
  if (nl == nr) {
    tree2[node] = ele;
    return;
  }
  long long mid = (nl + nr) / 2;
  if (nl <= ind && ind <= mid)
    update2(2 * node + 1, nl, mid, ind, ele, tree);
  else
    update2(2 * node + 2, mid + 1, nr, ind, ele, tree);
  tree2[node] = min(tree2[2 * node + 1], tree2[2 * node + 2]);
}
long long querry2(long long node, long long nl, long long nr, long long l,
                  long long r, vector<long long> &tree) {
  if (l > r || nl > nr || l > nr || nl > r) return 1000000000000000;
  if (l <= nl && nr <= r) return tree2[node];
  long long mid = (nl + nr) / 2;
  long long p1 = querry2(2 * node + 1, nl, mid, l, r, tree);
  long long p2 = querry2(2 * node + 2, mid + 1, nr, l, r, tree);
  return min(p1, p2);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    freq.clear();
    long long n;
    cin >> n;
    long long maxmon = 0;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      maxmon = max(maxmon, a[i]);
    }
    long long m;
    cin >> m;
    vector<pair<long long, long long> > b(m);
    long long maxhero = 0;
    for (int i = 0; i < m; i++) {
      cin >> b[i].first >> b[i].second;
      maxhero = max(maxhero, b[i].first);
    }
    if (maxhero < maxmon) {
      cout << -1 << '\n';
      continue;
    }
    for (int i = 0; i <= 4 * n; i++) {
      tree[i] = 0;
      tree2[i] = 1000000000000000;
    }
    sort(b.begin(), b.end());
    reverse(b.begin(), b.end());
    long long maxi = 0;
    for (int i = 0; i < m; i++) {
      maxi = max(maxi, b[i].second);
      freq[b[i].first] = maxi;
    }
    a.push_back(0);
    for (int i = 0; i <= n; i++) update(0, 0, n, i, a[i], tree);
    vector<long long> dp(n + 1, 0);
    update2(0, 0, n, n, 0, tree2);
    for (int i = n - 1; i >= 0; i--) {
      long long low = i, high = n;
      long long ans = -1;
      while (low <= high) {
        long long mid = (low + high) / 2;
        long long maxi = querry(0, 0, n, i, mid, tree);
        auto it = freq.lower_bound(maxi);
        long long endu = it->second;
        if (endu >= (mid - i + 1)) {
          ans = mid;
          low = mid + 1;
        } else {
          high = mid - 1;
        }
      }
      dp[i] = querry2(0, 0, n, i + 1, ans + 1, tree2) + 1;
      update2(0, 0, n, i, dp[i], tree2);
    }
    cout << dp[0] << '\n';
  }
  return 0;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0 || n < 0) return 0;
  long long ans = fact[n];
  long long temp = (fact[n - r] * fact[r]) % 1000000007;
  ans = (ans * inverse(temp, 1000000007)) % 1000000007;
  return ans;
}
void fileio() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
void build() {
  fact[0] = 1;
  for (long long i = 1; i < nax; i++) fact[i] = (fact[i - 1] * i) % 1000000007;
}
long long modulo(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent > 0) {
    if (exponent % 2 == 1) {
      result = (result * base) % modulus;
    }
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
void EE(long long a, long long b, long long &co1, long long &co2) {
  if (a % b == 0) {
    co1 = 0;
    co2 = 1;
    return;
  }
  EE(b, a % b, co1, co2);
  long long temp = co1;
  co1 = co2;
  co2 = temp - co2 * (a / b);
}
long long inverse(long long a, long long m) {
  long long x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
