#include <bits/stdc++.h>
using namespace std;
int xx[] = {0, 0, -1, 1}, yy[] = {-1, 1, 0, 0};
using ll = long long;
using ppi = pair<int, int>;
using ppl = pair<ll, ll>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int N;
    cin >> N;
    vector<int> hs(N), ss(N);
    for (long long i = (0); i < (N); ++i) {
      cin >> hs[i];
    }
    for (long long i = (0); i < (N); ++i) {
      cin >> ss[i];
    }
    const int originalH0 = hs[0];
    int l = 0, r = hs[0];
    hs.push_back(0);
    bool finalAns = false;
    while (l <= r) {
      int mid = (l + r) >> 1;
      hs[0] = mid;
      hs[N] = originalH0 - mid;
      int tag = 0;
      int capa = 0;
      for (long long i = (0); i < (N); ++i) {
        if (hs[i] > capa + ss[i]) {
          tag = 1;
          break;
        }
        capa = ss[i] - max(0, hs[i] - capa);
      }
      if (tag == 0 && capa < hs[N]) {
        tag = 2;
      }
      if (tag == 0) {
        finalAns = true;
        break;
      }
      if (tag == 1) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    cout << (finalAns ? "YES" : "NO") << ((char)10);
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 xx[] = {0, 0, -1, 1}, yy[] = {-1, 1, 0, 0};
using ll = long long;
using ppi = pair<int, int>;
using ppl = pair<ll, ll>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int N;
    cin >> N;
    vector<int> hs(N), ss(N);
    for (long long i = (0); i < (N); ++i) {
      cin >> hs[i];
    }
    for (long long i = (0); i < (N); ++i) {
      cin >> ss[i];
    }
    const int originalH0 = hs[0];
    int l = 0, r = hs[0];
    hs.push_back(0);
    bool finalAns = false;
    while (l <= r) {
      int mid = (l + r) >> 1;
      hs[0] = mid;
      hs[N] = originalH0 - mid;
      int tag = 0;
      int capa = 0;
      for (long long i = (0); i < (N); ++i) {
        if (hs[i] > capa + ss[i]) {
          tag = 1;
          break;
        }
        capa = ss[i] - max(0, hs[i] - capa);
      }
      if (tag == 0 && capa < hs[N]) {
        tag = 2;
      }
      if (tag == 0) {
        finalAns = true;
        break;
      }
      if (tag == 1) {
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    cout << (finalAns ? "YES" : "NO") << ((char)10);
  }
  return 0;
}

```