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 mod = 1000000007;
long long int INF = 1e18;
long long int po(long long int a, long long int b) {
  long long int re = 1;
  while (b) {
    if (b % 2 == 1) {
      re = (re * a);
      re %= mod;
    }
    a = (a * a);
    a %= mod;
    b >>= 1;
  }
  return re;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    bool g = 0;
    long long int a[n], b[n];
    for (int i = (0); i < (n); i++) cin >> a[i];
    for (int i = (0); i < (n); i++) cin >> b[i];
    long long int l = 0, r = b[0];
    while (l <= r) {
      bool f = 0;
      long long int mid = (l + r) / 2;
      long long int o = mid;
      for (int i = (0); i < (n); i++) {
        if (o > b[i]) {
          f = 1;
          break;
        }
        o = a[(i + 1) % n] - b[i] + o;
        if (o < 0) o = 0;
      }
      if (l == r && (f || o > mid)) break;
      if (f) {
        r = mid;
      } else if (o > mid) {
        l = mid + 1;
      } else {
        g = 1;
        break;
      }
    }
    if (g)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
