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;
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n), b(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i]);
  }
  auto calc = [&](int x) {
    for (int i = 0; i < n - 1; ++i) {
      int rem = max(a[i] - x, 0);
      x = b[i] - rem;
      if (x < 0) return -1;
    }
    return x;
  };
  int low = 0, hi = b[n - 1];
  while (low < hi) {
    int mid = (low + hi) >> 1;
    int res = calc(mid);
    if (res == -1 || (res < calc(mid + 1)))
      low = mid + 1;
    else
      hi = mid;
  }
  int x = calc(low);
  puts((x < 0 || x + (b[n - 1] - low) < a[n - 1]) ? "NO" : "YES");
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}
