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;
long long n, a[1000005], b[1000005], c[1000005], T, l, r;
bool answered;
bool check(long long now) {
  for (long long i = 1; i <= n; ++i) {
    c[i] = a[i];
  }
  c[1] -= now;
  for (long long i = 1; i < n; ++i) {
    if (c[i] > b[i]) {
      return true;
    }
    c[i + 1] -= b[i] - c[i];
    c[i + 1] = max(0ll, c[i + 1]);
  }
  if (b[n] - now < c[n]) {
    return false;
  }
  cout << "YES" << endl;
  answered = true;
  return false;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    answered = false;
    cin >> n;
    for (long long i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    for (long long i = 1; i <= n; ++i) {
      cin >> b[i];
    }
    l = 0, r = min(b[n], a[1]);
    while (l < r && !answered) {
      long long mid = (l + r) >> 1;
      if (check(mid)) {
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    if (!answered) {
      check(l);
      if (!answered) {
        cout << "NO" << endl;
      }
    }
  }
  return 0;
}
