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;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) ans = (ans << 3) + (ans << 1) + (c ^ 48);
  return ans * f;
}
inline int cmin(int &a, int b) { return a = min(a, b); }
inline int cmax(int &a, int b) { return a = max(a, b); }
int n;
int a[1000005], b[1000005], c[1000005];
int check(int mid) {
  for (int i = 1; i <= n; i++) c[i] = a[i];
  int lft = b[1] - mid;
  c[1] = a[1] - mid;
  for (int i = 2; i <= n; i++) {
    c[i] = max(c[i] - lft, 0);
    if (c[i] > b[i]) return 1;
    lft = b[i] - c[i];
    c[i] = 0;
  }
  if (c[1] > lft) return -1;
  return 0;
}
int main() {
  int T = read();
  while (T--) {
    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 = min(a[1], b[1]), flag = 0;
    while (l <= r) {
      int mid = l + r >> 1, ans = check(mid);
      if (ans == 0) {
        flag = 1;
        break;
      }
      if (ans < 0)
        l = mid + 1;
      else
        r = mid - 1;
    }
    if (flag)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
