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>
using namespace std;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  long long idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ", ";
  dbs(str.substr(idx + 1), s...);
}
template <class T>
void prc(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
template <class T>
void prall(T a) {
  prc(a.begin(), a.end());
}
const int inf = 1000 * 1000 * 1000 + 9;
const long long infl = 1ll * inf * inf;
int MOD = 1000 * 1000 * 1000 + 7;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
long long randrange(long long l, long long r) { return l + mt() % (r - l + 1); }
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  for (int i = 0; i < n; i++) cin >> a[i], a.push_back(a[i]);
  for (int i = 0; i < n; i++) cin >> b[i], b.push_back(b[i]);
  vector<int> done(n + n);
  for (int i = 0; i < n; i++) {
    if (done[i]) continue;
    done[i] = 1;
    int buf = 0;
    bool ok = 1;
    for (int j = 1; j < n; j++) {
      if (a[i + j] > buf + b[i + j]) {
        ok = 0;
        break;
      } else {
        done[i + j] = 1;
        if (a[i + j] <= buf)
          buf = b[i + j];
        else
          buf = b[i + j] + buf - a[i + j];
      }
    }
    if (ok && buf + b[i] >= a[i]) {
      cout << "YES" << '\n';
      return;
    }
  }
  cout << "NO" << '\n';
}
int main() {
  clock_t beg = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int testcases = 1;
  cin >> testcases;
  cout << fixed << setprecision(13);
  int cs = 1;
  while (testcases--) {
    cs++;
    solve();
  }
  clock_t end = clock();
  fprintf(stderr, "%lf sec\n", (double)(end - beg) / (CLOCKS_PER_SEC));
}
