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 = 1001000;
int n;
int a[N], A[N];
int b[N], B[N];
int Cut(int& x, int& y) {
  int Can = min(x, y);
  x -= Can;
  y -= Can;
}
bool Check(int Mid) {
  for (int i = 0; i < n; i++) A[i] = a[i];
  for (int i = 0; i < n; i++) B[i] = b[i];
  A[0] -= Mid;
  B[0] -= Mid;
  for (int i = 1; i < n; i++) {
    Cut(B[i - 1], A[i]);
    Cut(B[i], A[i]);
    if (A[i] > 0) return false;
  }
  return true;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", a + i);
    for (int i = 0; i < n; i++) scanf("%d", b + i);
    int Max = -1;
    int Low = 0, High = min(a[0], b[0]);
    while (Low <= High) {
      int Mid = Low + (High - Low) / 2;
      if (Check(Mid)) {
        Max = Mid;
        Low = Mid + 1;
      } else {
        High = Mid - 1;
      }
    }
    if (Max == -1) {
      puts("NO");
      continue;
    }
    Check(Max);
    Cut(B[n - 1], A[0]);
    Cut(B[n - 1], A[n - 1]);
    if (A[0] > 0 || A[n - 1] > 0) {
      puts("NO");
      continue;
    }
    puts("YES");
  }
}
