#include <bits/stdc++.h>
using namespace std;
template <typename T>
void deb(initializer_list<T> l) {
  for (auto& e : l) cout << e << ' ';
  cout << endl;
}
long long n;
pair<bool, long long> check(long long give, vector<long long> a,
                            vector<long long> b) {
  b[0] -= give;
  a[1] -= give;
  for (long long i = 1; i < n; ++i) {
    if (a[i] > b[i]) return {false, 0};
    b[i] -= a[i];
    a[i] = 0;
    a[(i + 1) % n] -= min(a[(i + 1) % n], b[i]);
  }
  a[0] -= min(a[0], b[0]);
  return {true, a[0]};
}
void solve() {
  cin >> n;
  vector<long long> a(n), b(n);
  for (auto& e : a) cin >> e;
  for (auto& e : b) cin >> e;
  long long low = 0;
  long long high = min(a[1], b[0]);
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    if (auto [poss, demand] = check(mid, a, b); poss) {
      if (demand == 0) {
        cout << "YES\n";
        return;
      }
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void deb(initializer_list<T> l) {
  for (auto& e : l) cout << e << ' ';
  cout << endl;
}
long long n;
pair<bool, long long> check(long long give, vector<long long> a,
                            vector<long long> b) {
  b[0] -= give;
  a[1] -= give;
  for (long long i = 1; i < n; ++i) {
    if (a[i] > b[i]) return {false, 0};
    b[i] -= a[i];
    a[i] = 0;
    a[(i + 1) % n] -= min(a[(i + 1) % n], b[i]);
  }
  a[0] -= min(a[0], b[0]);
  return {true, a[0]};
}
void solve() {
  cin >> n;
  vector<long long> a(n), b(n);
  for (auto& e : a) cin >> e;
  for (auto& e : b) cin >> e;
  long long low = 0;
  long long high = min(a[1], b[0]);
  while (low <= high) {
    long long mid = low + (high - low) / 2;
    if (auto [poss, demand] = check(mid, a, b); poss) {
      if (demand == 0) {
        cout << "YES\n";
        return;
      }
      high = mid - 1;
    } else
      low = mid + 1;
  }
  cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```