The government of Berland decided to improve network coverage in his country. Berland has a unique structure: the capital in the center and n cities in a circle around the capital. The capital already has a good network coverage (so the government ignores it), but the i-th city contains a_i households that require a connection.

The government designed a plan to build n network stations between all pairs of neighboring cities which will maintain connections only for these cities. In other words, the i-th network station will provide service only for the i-th and the (i + 1)-th city (the n-th station is connected to the n-th and the 1-st city).

All network stations have capacities: the i-th station can provide the connection to at most b_i households.

Now the government asks you to check can the designed stations meet the needs of all cities or not — that is, is it possible to assign each household a network station so that each network station i provides the connection to at most b_i households.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains the single integer n (2 ≤ n ≤ 10^6) — the number of cities and stations.

The second line of each test case contains n integers (1 ≤ a_i ≤ 10^9) — the number of households in the i-th city.

The third line of each test case contains n integers (1 ≤ b_i ≤ 10^9) — the capacities of the designed stations.

It's guaranteed that the sum of n over test cases doesn't exceed 10^6.

Output

For each test case, print YES, if the designed stations can meet the needs of all cities, or NO otherwise (case insensitive).

Example

Input


5
3
2 3 4
3 3 3
3
3 3 3
2 3 4
4
2 3 4 5
3 7 2 2
4
4 5 2 3
2 3 2 7
2
1 1
10 10


Output


YES
YES
NO
YES
YES

Note

In the first test case: 

  * the first network station can provide 2 connections to the first city and 1 connection to the second city; 
  * the second station can provide 2 connections to the second city and 1 connection to the third city; 
  * the third station can provide 3 connections to the third city. 



In the second test case: 

  * the 1-st station can provide 2 connections to the 1-st city; 
  * the 2-nd station can provide 3 connections to the 2-nd city; 
  * the 3-rd station can provide 3 connections to the 3-rd city and 1 connection to the 1-st station. 



In the third test case, the fourth city needs 5 connections, but the third and the fourth station has 4 connections in total.
#include <bits/stdc++.h>
void turn_off_de_bug() { freopen("/dev/null", "w", stderr); }
template <typename T, typename S>
std::ostream& operator<<(std::ostream& out, const std::pair<T, S>& p) {
  out << '(' << p.first << ", " << p.second << ")";
  return out;
}
template <typename T>
std::ostream& operator<<(std::ostream& out, const std::vector<T>& v) {
  if (!v.empty()) {
    out << '[';
    std::copy(v.begin(), v.end(), std::ostream_iterator<T>(out, ", "));
    out << "\b\b]";
  }
  return out;
}
template <typename T, typename S>
std::ostream& operator<<(std::ostream& out,
                         const std::set<std::pair<T, S> >& v) {
  out << '{';
  for (const auto& p : v) out << p << ", ";
  out << "\b\b}";
  return out;
}
template <typename T, typename S>
std::ostream& operator<<(std::ostream& out,
                         const std::multiset<std::pair<T, S> >& v) {
  out << '{';
  for (const auto& p : v) out << p << ", ";
  out << "\b\b}";
  return out;
}
template <typename T, typename S>
std::ostream& operator<<(std::ostream& out,
                         const std::vector<std::pair<T, S> >& v) {
  out << '[';
  for (const auto& p : v) out << p << ", ";
  out << "\b\b]";
  return out;
}
template <typename T, typename S>
std::ostream& operator<<(std::ostream& out,
                         const std::vector<std::vector<std::pair<T, S> > >& v) {
  out << '[';
  for (int i = 0; i < static_cast<int>(v.size()); ++i)
    out << '(' << i << ", " << v[i] << "), ";
  out << "\b\b]";
  return out;
}
const char* HR =
    "--------------------"
    "--------------------"
    "--------------------"
    "--------------------"
    "--------------------";
const int INF32 = int(1e9);
const long long INF64 = int(1e18);
int n;
std::vector<int> a, b;
void read() {
  std::cin >> n;
  a.resize(n);
  for (int& it : a) std::cin >> it;
  b.resize(n);
  for (int& it : b) std::cin >> it;
}
void solve() {
  long long need = 0;
  long long add = 0;
  long long extra = 2e9;
  for (int i = n - 1; i >= 0; --i) {
    if (b[i] < need) {
      std::cout << "NO\n";
      return;
    }
    long long val = b[i] - a[i];
    long long to_add = std::min(extra, std::max(0LL, val - need));
    add += to_add;
    extra = std::min(extra - to_add, b[i] - need - to_add);
    need = std::max(0LL, need - val);
  }
  std::cout << (add >= need ? "YES" : "NO");
  std::cout << "\n";
}
int main(int argc, char const* argv[]) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  turn_off_de_bug();
  int NUM_CASES = 1;
  std::cin >> NUM_CASES;
  while (NUM_CASES--) {
    read();
    solve();
  }
  return 0;
}
