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>
template <typename Tp>
void read(Tp &res) {
  static char ch;
  ch = getchar(), res = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
}
const int MAXN = 1e6 + 19;
int n, a[MAXN], b[MAXN], c[MAXN];
void check(int resev) {
  c[1] = resev, c[2] = b[1] - resev;
  for (int i = 3; i <= n; ++i) c[i] = 0;
  for (int i = 2; i <= n; ++i)
    if (c[i] >= a[i])
      c[i % n + 1] += b[i];
    else if (b[i] + c[i] >= a[i])
      c[i % n + 1] += b[i] - (a[i] - c[i]), c[i] = a[i];
    else
      c[i] += b[i];
}
int main() {
  int T;
  read(T);
  while (T--) {
    read(n);
    for (int i = 1; i <= n; ++i) read(a[i]);
    for (int i = 1; i <= n; ++i) read(b[i]);
    int L = 0, R = b[1];
    while (L < R) {
      int mid = (L + R) >> 1;
      check(mid);
      if (c[1] < a[1])
        L = mid + 1;
      else
        R = mid;
    }
    check(L);
    bool ans = 1;
    for (int i = 1; i <= n; ++i)
      if (c[i] < a[i]) ans = 0;
    puts(ans ? "YES" : "NO");
  }
  return 0;
}
