#include <bits/stdc++.h>
using namespace std;
int T, N, a[1000001], b[1000001];
bool chk[1000001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> N;
    for (int i = 0; i < N; i++) {
      cin >> a[i];
      chk[i] = 0;
    }
    for (int i = 0; i < N; i++) cin >> b[i];
    int s, prev = -1, cnt = 0;
    while (1) {
      int f = 0;
      for (s = 0; s < N; s++) {
        if (a[s] > b[s] && f == 0)
          f = 1;
        else if (f == 1 && b[s] >= a[s])
          break;
      }
      if (f == 0) {
        cout << "YES\n";
        break;
      } else if (s == N && a[0] > b[0]) {
        cout << "NO\n";
        break;
      }
      if (f == 1 && s == N) {
        s = 0;
      }
      chk[s] = 1;
      for (int i = 0; i < N; i++) {
        int now = (s + i) % N;
        int nxt = (now + 1) % N;
        if (b[now] < a[now]) continue;
        chk[nxt] = 0;
        int x = min(a[nxt], b[now] - a[now]);
        a[nxt] -= x;
        b[now] -= x;
      }
      cnt++;
      if (cnt == 31) {
        cout << "NO\n";
        break;
      }
    }
  }
}

### Prompt

Your task is to create 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;
int T, N, a[1000001], b[1000001];
bool chk[1000001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> T;
  while (T--) {
    cin >> N;
    for (int i = 0; i < N; i++) {
      cin >> a[i];
      chk[i] = 0;
    }
    for (int i = 0; i < N; i++) cin >> b[i];
    int s, prev = -1, cnt = 0;
    while (1) {
      int f = 0;
      for (s = 0; s < N; s++) {
        if (a[s] > b[s] && f == 0)
          f = 1;
        else if (f == 1 && b[s] >= a[s])
          break;
      }
      if (f == 0) {
        cout << "YES\n";
        break;
      } else if (s == N && a[0] > b[0]) {
        cout << "NO\n";
        break;
      }
      if (f == 1 && s == N) {
        s = 0;
      }
      chk[s] = 1;
      for (int i = 0; i < N; i++) {
        int now = (s + i) % N;
        int nxt = (now + 1) % N;
        if (b[now] < a[now]) continue;
        chk[nxt] = 0;
        int x = min(a[nxt], b[now] - a[now]);
        a[nxt] -= x;
        b[now] -= x;
      }
      cnt++;
      if (cnt == 31) {
        cout << "NO\n";
        break;
      }
    }
  }
}

```