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;
const int N = 2e6 + 9, Mod = 1e9 + 7;
int n, q, m, A[N], B[N];
long long PA[N], PB[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  scanf("%d", &q);
  for (; q; q--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &A[i]), A[i + n] = A[i];
    for (int i = 0; i < n; i++)
      scanf("%d", &B[i]), B[i] = min(B[i], A[i] + A[i + 1]), B[i + n] = B[i];
    for (int i = 1; i <= n + n; i++)
      PA[i] = PA[i - 1] + A[i - 1], PB[i] = PB[i - 1] + B[i - 1];
    bool Fail = 0;
    if (PB[n] < PA[n]) Fail = 1;
    priority_queue<long long> Pq, Qp;
    for (int i = 1; i <= n + n && !Fail; i++) {
      long long rval = PB[i] - PA[i];
      while (Pq.size() && Qp.size() && Pq.top() == Qp.top()) Pq.pop(), Qp.pop();
      if (Pq.size() && Pq.top() > rval) {
        Fail = 1;
        break;
      }
      long long lval = PB[i - 1] - PA[i];
      Pq.push(lval);
      if (i >= n - 1) {
        int j = i - n + 2;
        long long llval = PB[j - 1] - PA[j];
        Qp.push(llval);
      }
    }
    if (Fail) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
  }
  return 0;
}
