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;
int N;
int A[1000000 + 4];
int B[1000000 + 4];
bool check(int x0) {
  int rem_cap = B[0] - x0;
  for (int i = (1), _b = (N - 1); i <= _b; ++i) {
    int m = min(A[i], rem_cap);
    int rem_houses = A[i] - m;
    rem_cap = B[i] - rem_houses;
    if (rem_cap < 0) return false;
  }
  return true;
}
bool check2(int x0, int x1) {
  int rem_cap = B[0] - x0;
  for (int i = (1), _b = (N - 1); i <= _b; ++i) {
    int m = min(A[i], rem_cap);
    int rem_houses = A[i] - m;
    rem_cap = B[i] - rem_houses;
    assert(rem_cap >= 0);
  }
  return rem_cap >= x1;
}
bool solve() {
  int lo = 0, hi = min(A[0], B[0]);
  int min0 = -1;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    if (check(mid)) {
      lo = mid + 1;
      min0 = mid;
    } else {
      hi = mid - 1;
    }
  }
  if (min0 < 0) return false;
  return check2(min0, A[0] - min0);
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int TC;
  cin >> TC;
  for (int tc = (1), _b = (TC); tc <= _b; ++tc) {
    cin >> N;
    for (int i = 0, _n = (N); i < _n; ++i) cin >> A[i];
    for (int i = 0, _n = (N); i < _n; ++i) cin >> B[i];
    bool res = solve();
    cout << (res ? "YES" : "NO") << '\n';
  }
  return 0;
}
