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 power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long a[n];
    long long b[n];
    long long c[n], d[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      c[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      d[i] = b[i];
    }
    int check = 1;
    c[0] = min(d[n - 1], a[0]);
    d[n - 1] -= c[0];
    long long p[n];
    p[0] = c[0];
    for (int i = 0; i < n - 1; i++) {
      if (i != 0) {
        c[i] += min(d[(n + i - 1) % n], a[i] - c[i]);
        d[(n + i - 1) % n] -= c[i];
        p[i] = c[i];
      }
      long long v1 = min(d[i], a[i] - c[i]);
      d[i] -= v1;
      c[i] += v1;
      if (c[i] < a[i]) {
        check = 0;
        break;
      }
    }
    if (check == 0) {
      cout << "NO\n";
    } else {
      long long val = 0ll;
      for (int i = n - 3; i >= 0; i--) {
        if (d[i] > 0) {
          val = min(d[i], min(b[(n + i - 1) % n] - d[(n + i - 1) % n], p[i]));
          d[(n + i - 1) % n] += val;
        }
      }
      if (d[n - 2] + d[n - 1] >= a[n - 1])
        cout << "YES\n";
      else
        cout << "NO\n";
    }
  }
  return 0;
}
