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 int power(long long int x, unsigned long long int y,
                    long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    for (long long int k_ = 0; k_ < n; k_++) {
      cin >> a[k_];
    };
    vector<long long int> b(n);
    for (long long int k_ = 0; k_ < n; k_++) {
      cin >> b[k_];
    };
    long long int i = 0, empty_i = 0;
    vector<pair<long long int, long long int> > bridge(n);
    long long int prevb = 0;
    long long int flag = 0;
    while (1) {
      i++;
      i %= n;
      if (prevb >= a[i])
        prevb = b[i];
      else if (prevb + b[i] >= a[i]) {
        prevb = prevb + b[i] - a[i];
      } else {
        empty_i = i;
        prevb = 0;
        flag = 1;
      }
      if (i == 0) break;
    }
    if (empty_i == 0 and flag == 1) {
      cout << "No\n";
      continue;
    }
    i = empty_i;
    long long int flag2 = 0;
    prevb = 0;
    while (1) {
      i++;
      i %= n;
      if (prevb >= a[i])
        prevb = b[i];
      else if (prevb + b[i] >= a[i]) {
        prevb = prevb + b[i] - a[i];
      } else {
        flag2 = 1;
        break;
      }
      if (i == empty_i) break;
    }
    if (flag2)
      cout << "No\n";
    else
      cout << "Yes\n";
  }
  return 0;
}
