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 long long inf = 0x3f3f3f3f3f3f3f3f;
const long long N = 1000000;
long long n;
long long a[N + 1], b[N + 1];
long long Suma[N + 1], Sumb[N + 1];
void mian() {
  cin >> n;
  for (long long i = 1; i <= n; i++)
    scanf("%lld", a + i), Suma[i] = Suma[i - 1] + a[i];
  for (long long i = 1; i <= n; i++)
    scanf("%lld", b + i), Sumb[i] = Sumb[i - 1] + b[i];
  if (Sumb[n] - Suma[n] < 0) return puts("NO"), void();
  long long mn = inf;
  for (long long i = 1; i <= n; i++) {
    if (i > 1) mn = min(mn, -Sumb[i - 1] + Suma[i]);
    if (Sumb[i] - Suma[i] + mn < 0) return puts("NO"), void();
  }
  long long mx = 0;
  mn = inf;
  for (long long i = 2; i <= n; i++)
    mx = max(mx, Suma[n] - Suma[i] + a[1] - Sumb[n] + Sumb[i - 1]);
  for (long long i = 1; i <= n; i++) mn = min(mn, Sumb[i] - Suma[i] + a[1]);
  puts(mx <= mn ? "YES" : "NO");
}
signed main() {
  long long testnum;
  cin >> testnum;
  while (testnum--) mian();
  return 0;
}
