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>
const int N = 2e6 + 5;
using namespace std;
int T, n, pos, a[N], b[N], c[N], flag;
long long mx, sum1, sum2;
template <typename T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * w;
}
int main() {
  T = read<int>();
  while (T--) {
    n = read<int>(), flag = 1, sum1 = sum2 = 0;
    for (int i = 1; i <= n; i++) a[i] = a[i + n] = read<int>();
    for (int i = 1; i <= n; i++) b[i] = b[i + n] = read<int>();
    for (int i = 1; i <= n; i++) sum1 += a[i], sum2 += b[i];
    if (sum2 < sum1) {
      puts("NO");
      continue;
    }
    mx = 0, sum1 = b[n], sum2 = 0;
    for (int i = 1; i <= 2 * n; i++) {
      sum1 += b[i], sum2 += a[i];
      if (sum1 - sum2 < mx) {
        flag = 0;
        break;
      }
      if ((sum1 - b[i]) - sum2 > mx) mx = sum1 - b[i] - sum2;
    }
    puts(flag ? "YES" : "NO");
  }
  return 0;
}
