#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 50;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
int n;
long long a[MAXN], b[MAXN];
long long sa[MAXN], sb[MAXN];
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i), sa[i] = a[i];
  for (int i = 1; i <= n; i++) scanf("%lld", b + i), sb[i] = b[i];
  long long x = 0, now = 0;
  for (int i = 2; i <= n; i++) now += b[i - 1] - a[i], x = max(x, now);
  x = min(b[1], x);
  bool ok = 1;
  a[1] -= x;
  b[1] -= x;
  for (int i = 2; i <= n; i++) {
    int tmp = min(a[i], b[i - 1]);
    a[i] -= tmp;
    b[i - 1] -= tmp;
    tmp = min(a[i], b[i]);
    a[i] -= tmp;
    b[i] -= tmp;
    if (a[i] > 0) {
      x -= a[i];
    }
  }
  if (x < 0) ok = 0;
  for (int i = 1; i <= n; i++) a[i] = sa[i], b[i] = sb[i];
  a[1] -= x;
  b[1] -= x;
  for (int i = 2; i <= n; i++) {
    int tmp = min(a[i], b[i - 1]);
    a[i] -= tmp;
    b[i - 1] -= tmp;
    tmp = min(a[i], b[i]);
    a[i] -= tmp;
    b[i] -= tmp;
    if (a[i] > 0) {
      ok = 0;
      break;
    }
  }
  if (a[1] > b[n]) ok = 0;
  printf("%s\n", ok ? "YES" : "NO");
}
int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; cas++) work();
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 50;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
int n;
long long a[MAXN], b[MAXN];
long long sa[MAXN], sb[MAXN];
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i), sa[i] = a[i];
  for (int i = 1; i <= n; i++) scanf("%lld", b + i), sb[i] = b[i];
  long long x = 0, now = 0;
  for (int i = 2; i <= n; i++) now += b[i - 1] - a[i], x = max(x, now);
  x = min(b[1], x);
  bool ok = 1;
  a[1] -= x;
  b[1] -= x;
  for (int i = 2; i <= n; i++) {
    int tmp = min(a[i], b[i - 1]);
    a[i] -= tmp;
    b[i - 1] -= tmp;
    tmp = min(a[i], b[i]);
    a[i] -= tmp;
    b[i] -= tmp;
    if (a[i] > 0) {
      x -= a[i];
    }
  }
  if (x < 0) ok = 0;
  for (int i = 1; i <= n; i++) a[i] = sa[i], b[i] = sb[i];
  a[1] -= x;
  b[1] -= x;
  for (int i = 2; i <= n; i++) {
    int tmp = min(a[i], b[i - 1]);
    a[i] -= tmp;
    b[i - 1] -= tmp;
    tmp = min(a[i], b[i]);
    a[i] -= tmp;
    b[i] -= tmp;
    if (a[i] > 0) {
      ok = 0;
      break;
    }
  }
  if (a[1] > b[n]) ok = 0;
  printf("%s\n", ok ? "YES" : "NO");
}
int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; cas++) work();
}

```