#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
int n;
vector<int> a, b;
long long get(long long prev) {
  for (int i = 1; i < n; ++i) {
    long long get = max(0LL, a[i] - prev);
    if (get > b[i]) return -1;
    prev = b[i] - get;
  }
  return prev;
}
void solve() {
  read(n);
  a.resize(n);
  b.resize(n);
  for (int i = 0; i < n; ++i) {
    read(a[i]);
  }
  for (int i = 0; i < n; ++i) {
    read(b[i]);
  }
  int st = 0, en = b[0];
  if (get(en) == -1) {
    prn("NO");
    return;
  }
  while (st < en) {
    int mid = (st + en) / 2;
    if (get(mid) != -1)
      en = mid;
    else
      st = mid + 1;
  }
  long long lft = b[0] - st;
  long long g = get(st), need = max(0LL, a[0] - lft);
  if (need <= g)
    prn("YES");
  else
    prn("NO");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  read(t);
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
template <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
int n;
vector<int> a, b;
long long get(long long prev) {
  for (int i = 1; i < n; ++i) {
    long long get = max(0LL, a[i] - prev);
    if (get > b[i]) return -1;
    prev = b[i] - get;
  }
  return prev;
}
void solve() {
  read(n);
  a.resize(n);
  b.resize(n);
  for (int i = 0; i < n; ++i) {
    read(a[i]);
  }
  for (int i = 0; i < n; ++i) {
    read(b[i]);
  }
  int st = 0, en = b[0];
  if (get(en) == -1) {
    prn("NO");
    return;
  }
  while (st < en) {
    int mid = (st + en) / 2;
    if (get(mid) != -1)
      en = mid;
    else
      st = mid + 1;
  }
  long long lft = b[0] - st;
  long long g = get(st), need = max(0LL, a[0] - lft);
  if (need <= g)
    prn("YES");
  else
    prn("NO");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  read(t);
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}

```