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>
#pragma optimize("-O3")
using namespace std;
const long long mod = 998244353;
int n, a[1000006], b[1000006];
int extra[1000006];
void no() { cout << "NO" << '\n'; }
void yes() { cout << "YES" << '\n'; }
void solve() {
  cin >> n;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) cin >> a[i];
  ;
  for (long long i = (long long)0; i != (long long)n; i = i + 1) cin >> b[i];
  ;
  extra[0] = b[0];
  for (long long i = (long long)1; i != (long long)n; i = i + 1) {
    extra[i] = max(0, min(b[i], extra[i - 1] + (b[i] - a[i])));
  }
  extra[0] = max(0, min(b[0], extra[n - 1] + (b[0] - a[0])));
  for (long long i = (long long)1; i != (long long)n; i = i + 1) {
    extra[i] = max(0, min(b[i], extra[i - 1] + (b[i] - a[i])));
  }
  for (long long i = (long long)0; i != (long long)n; i = i + 1) {
    if (extra[i] + b[(i + 1) % n] < a[(i + 1) % n] + extra[(i + 1) % n]) {
      no();
      return;
    }
  }
  yes();
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}
