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;
int32_t seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline int32_t rnd(int32_t l = 0, int32_t r = INT_MAX) {
  return uniform_int_distribution<int32_t>(l, r)(rng);
}
template <typename T, typename G>
ostream& operator<<(ostream& out, const pair<T, G>& a) {
  return out << "( " << a.first << ", " << a.second << ")";
}
template <typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
  out << "[ ";
  for (const auto& i : a) out << i << ", ";
  return out << "]" << endl;
}
template <typename T>
ostream& operator<<(ostream& out, const set<T>& a) {
  out << "{ ";
  for (const auto& i : a) out << i << ", ";
  return out << "}" << endl;
}
template <typename T, typename G>
ostream& operator<<(ostream& out, const map<T, G>& a) {
  out << "< ";
  for (const auto& i : a) out << i << ", ";
  return out << ">" << endl;
}
template <typename T, size_t N>
typename enable_if<!is_same<typename remove_cv<T>::type, char>::value,
                   ostream&>::type
operator<<(ostream& out, T (&a)[N]) {
  out << "[ ";
  for (size_t i = 0; i < N; ++i) out << a[i] << ", ";
  out << "]" << endl;
  return out;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << (H) << " ";
  debug_out(T...);
}
int32_t chk(int32_t start, vector<int32_t>& a, vector<int32_t>& b) {
  int32_t i, n = a.size();
  vector<int32_t> distribute(n, 0);
  distribute[0] = min(start, a[0]);
  int32_t past = b[0] - distribute[0];
  for (i = 1; i < n; i++) {
    int32_t rem = min(a[i], past);
    past = b[i];
    distribute[i] += rem;
    if (distribute[i] < a[i]) {
      rem = min(a[i] - distribute[i], past);
      distribute[i] += rem;
      past -= rem;
    }
  }
  if (a[0] - distribute[0] > 0)
    distribute[0] += min(a[0] - distribute[0], past);
  long long sm = 0;
  24;
  for (i = 0; i < n; i++) sm += max(a[i] - distribute[i], a[i] - a[i]);
  24;
  return sm;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int32_t t;
  cin >> t;
  while (t--) {
    int32_t n, i;
    cin >> n;
    vector<int32_t> a(n), b(n);
    for (i = 0; i < n; i++) cin >> a[i];
    for (i = 0; i < n; i++) cin >> b[i];
    int32_t l = 0, r = min(b[0], a[0]);
    while (r - l >= 4) {
      int32_t ml = l + (r - l) / 3;
      int32_t mr = r - (r - l) / 3;
      int32_t sl = chk(ml, a, b);
      int32_t sr = chk(mr, a, b);
      if (sl < sr)
        r = mr;
      else
        l = ml;
    }
    int32_t ans = n;
    for (i = l; i <= r; i++) {
      ans = min(ans, chk(i, a, b));
    }
    if (ans == 0)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
