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 <class T>
void Output(const T& v, bool print_new_line = true) {
  bool first = true;
  for (auto element : v) {
    if (!first)
      cout << " ";
    else
      first = false;
    cout << element;
  }
  if (print_new_line) cout << endl;
}
template <class T>
void Input(vector<T>& v, int n) {
  v.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
  }
}
template <class T>
vector<int> Query(const T& c, int answer_counts) {
  cout << "? " << c.size();
  for (auto it = c.begin(); it != c.end(); ++it) {
    cout << " " << (*it);
  }
  cout << endl;
  cout.flush();
  vector<int> s(answer_counts);
  for (int i = 0; i < answer_counts; ++i) {
    cin >> s[i];
  }
  return s;
}
int check(int c0, const vector<int>& city, const vector<int>& network) {
  int t = c0;
  for (int i = 0; i < city.size() - 1; ++i) {
    int rest = network[i] - t;
    if (rest < 0) return -1;
    t = city[i + 1] - rest;
    if (t < 0) t = 0;
  }
  if (t > network[city.size() - 1]) {
    return -1;
  }
  return network[city.size() - 1] - t;
}
int main() {
  int case_number;
  cin >> case_number;
  for (int case_index = 0; case_index < case_number; ++case_index) {
    int n;
    cin >> n;
    vector<int> city, network;
    Input(city, n);
    Input(network, n);
    int low = 0, up = min(city[0], network[0]);
    int last_t = check(low, city, network);
    if (last_t < 0) {
      cout << "NO" << endl;
      continue;
    }
    int t = check(up, city, network);
    if (t >= 0) {
      cout << (t + up >= city[0] ? "YES" : "NO") << endl;
      continue;
    }
    while (low < up - 1) {
      int center = (low + up) / 2;
      int t = check(center, city, network);
      if (t >= 0) {
        last_t = t;
        low = center;
      } else {
        up = center;
      }
    }
    cout << (low + last_t >= city[0] ? "YES" : "NO") << endl;
  }
  return 0;
}
