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;
int read() {
  int res = 0, ch, flag = 0;
  if ((ch = getchar()) == '-')
    flag = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return flag ? -res : res;
}
const int inf = 0x3f3f3f3f;
const int N = 1000000 + 10;
int a[N];
int b[N];
int a1[N];
int b1[N];
int n;
int sum;
bool check(int mid) {
  for (int i = 1; i <= n; i++) a1[i] = a[i], b1[i] = b[i];
  a1[1] = mid;
  for (int i = 1; i <= n; i++) {
    b1[i] -= a1[i];
    if (b1[i] < 0) return false;
    a1[i + 1] -= b1[i];
    a1[i + 1] = max(a1[i + 1], 0);
    if (i == n) sum = mid + b1[i];
  }
  return true;
}
void solve() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) b[i] = read();
  int l = 0, r = b[1] + 1;
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  if (!check(l)) {
    cout << "NO" << endl;
    return;
  }
  if (sum < a[1]) {
    cout << "NO" << endl;
    return;
  }
  cout << "YES" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
