#include <bits/stdc++.h>
using namespace std;
int check(vector<int>& A, vector<int>& B, int ans) {
  vector<int> AP = A;
  vector<int> BP = B;
  AP.push_back(A[0] - ans);
  AP[0] = ans;
  int carry = 0;
  for (int i = 0; i < BP.size(); i++) {
    if (BP[i] < AP[i]) return 1;
    BP[i] -= AP[i];
    AP[i] = 0;
    AP[i + 1] -= min(AP[i + 1], BP[i]);
  }
  if (*AP.rbegin() == 0) return 0;
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> A(n);
    vector<int> B(n);
    for (auto& Ai : A) cin >> Ai;
    for (auto& Bi : B) cin >> Bi;
    int high = A[0];
    int low = 0;
    while (low < high) {
      int mid = (low + high) / 2;
      int res = check(A, B, mid);
      if (res == 0) {
        low = high = mid;
        break;
      }
      if (res < 0) {
        low = mid + 1;
      }
      if (res > 0) {
        high = mid - 1;
      }
    }
    if (check(A, B, low) == 0) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int check(vector<int>& A, vector<int>& B, int ans) {
  vector<int> AP = A;
  vector<int> BP = B;
  AP.push_back(A[0] - ans);
  AP[0] = ans;
  int carry = 0;
  for (int i = 0; i < BP.size(); i++) {
    if (BP[i] < AP[i]) return 1;
    BP[i] -= AP[i];
    AP[i] = 0;
    AP[i + 1] -= min(AP[i + 1], BP[i]);
  }
  if (*AP.rbegin() == 0) return 0;
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> A(n);
    vector<int> B(n);
    for (auto& Ai : A) cin >> Ai;
    for (auto& Bi : B) cin >> Bi;
    int high = A[0];
    int low = 0;
    while (low < high) {
      int mid = (low + high) / 2;
      int res = check(A, B, mid);
      if (res == 0) {
        low = high = mid;
        break;
      }
      if (res < 0) {
        low = mid + 1;
      }
      if (res > 0) {
        high = mid - 1;
      }
    }
    if (check(A, B, low) == 0) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}

```