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;
template <typename T>
inline void read(T &x) {
  x = 0;
  int fu = 1;
  char c = getchar();
  while (c > 57 || c < 48) {
    if (c == 45) fu = -1;
    c = getchar();
  }
  while (c <= 57 && c >= 48) {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  x *= fu;
}
template <typename T>
inline void fprint(T x) {
  if (x < 0) {
    putchar(45);
    x = -x;
  }
  if (x > 9) fprint(x / 10);
  putchar(x % 10 + 48);
}
int n;
int a[1000010], b[1000010];
int c[1000010], d[1000010];
int check(int x) {
  for (register int i = 1; i <= n; i++) {
    c[i] = a[i];
    d[i] = b[i];
  }
  c[1] -= x;
  d[1] -= x;
  for (register int i = 2; i <= n; i++) {
    c[i] = max(0, c[i] - d[i - 1]);
    d[i] -= c[i];
    if (d[i] < 0) {
      return -1;
    }
  }
  if (c[1] > d[n])
    return 1;
  else
    return 0;
}
bool find(int l, int r) {
  while (r - l >= 5) {
    int mid = (l + r) >> 1;
    int ret = check(mid);
    if (ret == 0) return true;
    if (ret < 0) r = mid - 1;
    if (ret > 0) l = mid + 1;
  }
  for (register int i = l; i <= r; i++) {
    if (check(i) == 0) return true;
  }
  return false;
}
inline void work() {
  read(n);
  for (register int i = 1; i <= n; i++) read(a[i]);
  for (register int i = 1; i <= n; i++) read(b[i]);
  if (find(0, min(a[1], b[1]))) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
}
int main() {
  int _____;
  read(_____);
  while (_____--) {
    work();
  }
}
