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;
const int nax = 1e6 + 5;
int a[nax];
int b[nax];
long long int res[nax];
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  long long int x = 0;
  for (int i = 1; i <= n; i++) {
    x += b[i];
    x -= a[i];
  }
  if (x < 0) {
    cout << "NO"
         << "\n";
    return;
  }
  long long int akt = 0;
  multiset<long long int> s;
  for (int i = 2; i <= n; i++) {
    akt += a[i];
    akt -= b[i];
    res[i] = akt;
    s.insert(akt);
  }
  long long int v = *s.rbegin();
  if (v - b[1] > 0) {
    cout << "NO"
         << "\n";
    return;
  }
  long long int add = 0;
  long long int pp = 0;
  long long int xd = 0;
  for (int i = 2; i <= n; i++) {
    add += a[i - 1];
    add -= b[i - 1];
    xd -= a[i];
    xd += b[i];
    pp = max(pp, add);
    s.erase(s.find(res[i]));
    long long int val = 0;
    if (i < n) {
      long long int pot;
      pot = *s.rbegin() + xd;
      pot = max(pot, res[n] + xd + pp);
      if (pot - b[i] > 0) {
        cout << "NO"
             << "\n";
        return;
      }
    } else {
      long long int qq = -1e18;
      long long int wa = 0;
      for (int j = 1; j < n; j++) {
        wa += a[j];
        wa -= b[j];
        qq = max(wa, qq);
      }
      if (qq - b[n] > 0) {
        cout << "NO"
             << "\n";
        return;
      }
    }
  }
  cout << "YES"
       << "\n";
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
