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>
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
template <typename T>
void print_vec(vector<T> v) {
  for (int x : v) cout << x << ' ';
  cout << endl;
}
const long long INF = 1e15;
bool judge(vector<long long> a, vector<long long> b, long long x) {
  int n = a.size();
  b[0] -= x;
  a[1] -= x;
  for (int i = 1; i < n; i++) {
    if (i != 0) {
      if (a[i] > b[i]) {
        return false;
      }
      long long d = min(a[i], b[i]);
      b[i] -= d;
      a[i] -= d;
    }
    int nx = (i + 1) % n;
    long long d = min(b[i], a[nx]);
    a[nx] -= d;
    b[i] -= d;
  }
  return true;
}
bool ok(vector<long long> a, vector<long long> b, long long x) {
  int n = a.size();
  b[0] -= x;
  a[1] -= x;
  for (int i = 1; i < n; i++) {
    if (i != 0) {
      if (a[i] > b[i]) {
        return false;
      }
      long long d = min(a[i], b[i]);
      b[i] -= d;
      a[i] -= d;
    }
    int nx = (i + 1) % n;
    long long d = min(b[i], a[nx]);
    a[nx] -= d;
    b[i] -= d;
  }
  return a[0] <= b[0] + b[n - 1];
}
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n), b(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  int l = 0, r = min(a[1], b[0]);
  if (judge(a, b, l)) {
    r = 0;
  } else if (!judge(a, b, r)) {
    cout << "NO" << endl;
    return;
  } else {
    while (r - l > 1) {
      int c = (l + r) / 2;
      if (judge(a, b, c))
        r = c;
      else
        l = c;
    }
  }
  if (ok(a, b, r)) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(10) << fixed;
  int t;
  cin >> t;
  while (t--) solve();
}
