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>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T maxs(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T mins(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long solve() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1), b(n + 1);
  long long s1 = 0, s2 = 0;
  for (long long i = 1; i < n + 1; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i < n + 1; i++) {
    cin >> b[i];
  }
  auto f = [&](long long first) {
    long long val = a[1] - b[1] + first;
    if (val > b[n]) {
      return 1LL;
    }
    long long temp = b[n];
    b[n] = min(b[n], b[n] - val);
    long long flag = 0;
    for (long long i = 2; i <= n; i++) {
      long long val = a[i] - first;
      if (val > b[i]) {
        b[n] = temp;
        return flag;
      }
      if (val < 0) flag = 1;
      first = min(b[i], b[i] - val);
    }
    b[n] = temp;
    return 2LL;
  };
  long long l = 0, r = b[1];
  while (l <= r) {
    long long mid = (l + r) / 2;
    long long first = f(mid);
    if (first == 2) {
      cout << "YES"
           << "\n";
      return 0;
    } else {
      if (first == 1)
        r = mid - 1;
      else
        l = mid + 1;
    }
  }
  cout << "NO"
       << "\n";
  return 0;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
