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 check(vector<long long>& a, vector<long long>& b, long long c) {
  long long crprev = b[0] - c;
  for (long long i = 1; i < (long long)a.size(); ++i) {
    long long cl = max(a[i] - crprev, 0ll);
    if (cl > b[i]) {
      return -1;
    }
    crprev = b[i] - cl;
  }
  return (crprev + c) < a[0];
}
bool F(vector<long long>& a, vector<long long>& b) {
  long long lb = 0;
  long long ub = min(a[0], b[0]) + 1;
  while (lb + 1 < ub) {
    long long mid = (lb + ub) / 2;
    long long c = check(a, b, mid);
    if (c < 0) {
      ub = mid;
    } else if (c > 0) {
      lb = mid;
    } else {
      return true;
    }
  }
  return check(a, b, lb) == 0;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  for (; t-- > 0;) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (auto& ai : a) {
      cin >> ai;
    }
    vector<long long> b(n);
    for (auto& bi : b) {
      cin >> bi;
    }
    cout << (F(a, b) ? "YES" : "NO") << endl;
  }
  return 0;
}
