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;
using ii = pair<int, int>;
using lint = long long;
const int MOD = 1e9 + 7, mxn = 1000100;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
int tc, N, a[mxn], b[mxn], bb[mxn];
vector<vector<int>> adj;
lint tota, totb;
void input() {
  cin >> N;
  tota = 0, totb = 0;
  for (int i = 0; i < N; i++) cin >> a[i], tota += a[i];
  for (int i = 0; i < N; i++) cin >> b[i], totb += b[i], b[i];
}
int f(int res) {
  int cry = res;
  for (int i = 1; i < N; i++) {
    if (cry + b[i] - a[i] < 0) return -1;
    cry = min(b[i], cry + b[i] - a[i]);
  }
  return a[0] <= cry + (b[0] - res);
}
int solve() {
  if (tota > totb) return 0;
  int s = 0, e = b[0];
  for (int i = 0; i < N; i++)
    if (b[(i - 1 + N) % N] + b[i] < a[i]) return 0;
  while (s <= e) {
    int mid = (s + e) / 2;
    int x = f(mid);
    if (x == 1) return 1;
    if (!x)
      e = mid - 1;
    else
      s = mid + 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> tc;
  while (tc--) {
    input();
    int ret = solve();
    if (ret)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
