#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
int q;
int n;
long long ar[N], mx[N], pre[N];
long long tot, cur;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  while (q--) {
    bool ok = 1;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> ar[i];
      pre[i + i] = -ar[i];
    }
    for (int i = 1; i <= n; i++) {
      cin >> mx[i];
      int id = (i + i) % (n + n) + 1;
      pre[id] = mx[i];
    }
    n = n + n;
    for (int i = 1; i <= n; i++) {
      pre[n + i] = pre[i];
    }
    for (int i = 1; i <= n + n; i++) {
      pre[i] = pre[i] + pre[i - 1];
    }
    if (pre[n] < 0) {
      cout << "NO" << '\n';
      continue;
    }
    long long mx = 0;
    for (int i = 0; i <= n + n; i++) {
      if (i & 1) {
        if (pre[i] - mx < 0) {
          ok = 0;
          break;
        }
      } else {
        mx = max(mx, pre[i]);
      }
    }
    if (ok)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 N = 2e6 + 10;
int q;
int n;
long long ar[N], mx[N], pre[N];
long long tot, cur;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  while (q--) {
    bool ok = 1;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> ar[i];
      pre[i + i] = -ar[i];
    }
    for (int i = 1; i <= n; i++) {
      cin >> mx[i];
      int id = (i + i) % (n + n) + 1;
      pre[id] = mx[i];
    }
    n = n + n;
    for (int i = 1; i <= n; i++) {
      pre[n + i] = pre[i];
    }
    for (int i = 1; i <= n + n; i++) {
      pre[i] = pre[i] + pre[i - 1];
    }
    if (pre[n] < 0) {
      cout << "NO" << '\n';
      continue;
    }
    long long mx = 0;
    for (int i = 0; i <= n + n; i++) {
      if (i & 1) {
        if (pre[i] - mx < 0) {
          ok = 0;
          break;
        }
      } else {
        mx = max(mx, pre[i]);
      }
    }
    if (ok)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}

```