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>
namespace {
using namespace std;
template <typename T>
inline bool whitespace(const vector<T> &) {
  return false;
}
template <typename T>
inline bool whitespace(const T &) {
  return true;
}
inline bool whitespace(const char) { return false; }
template <typename F, typename S>
inline bool whitespace(const pair<F, S> &) {
  return false;
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  os << p.first << (whitespace(p.second) ? " " : "") << p.second << "\n";
  return os;
}
template <typename F, typename S>
istream &operator>>(istream &is, pair<F, S> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << (whitespace(t) ? " " : "");
  }
  os << "\n";
  return os;
}
template <typename T>
istream &operator>>(istream &is, deque<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &v) {
  for (const T &t : v) {
    os << t << (whitespace(t) ? " " : "");
  }
  os << "\n";
  return os;
}
class exit_exception : exception {};
void answer() {}
template <typename First, typename... Args>
void answer(First &&val, Args &&...ans) {
  cout << val << "\n";
  answer(ans...);
  throw exit_exception();
}
struct autoint {
  long long x;
  autoint(long long i) : x(i) {}
  autoint() : x(0) { cin >> x; }
  operator long long() { return x; }
  long long operator--() { return --x; }
  long long operator--(signed) { return x--; }
  long long operator++() { return ++x; }
  long long operator++(signed) { return x++; }
};
template <typename T>
T gcd(T a, T b) {
  while (b) {
    T mod = a % b;
    a = b;
    b = mod;
  }
  return a;
}
template <typename T>
vector<T> prefix_sum(const vector<T> &v, long long from = 0) {
  vector<T> res(v.size() + 1 - from);
  res[0] = 0;
  for (long long i = from; i < v.size(); ++i)
    res[i + 1 - from] = res[i - from] + v[i];
  return res;
}
}  // namespace
void solve();
signed main() {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  long long t = 1;
  cin >> t;
  do {
    try {
      solve();
    } catch (exit_exception &) {
    }
  } while (--t);
  return 0;
}
long long test(vector<long long> a, vector<long long> b) {
  for (long long i = 1; i < a.size(); ++i) {
    if (a[i] > b[i - 1])
      a[i] -= b[i - 1];
    else
      a[i] = 0;
    b[i] -= a[i];
    if (b[i] < 0) return -1;
  }
  return b.back();
}
vector<long long> add(long long i, long long x, vector<long long> v) {
  v[i] += x;
  return v;
}
void solve() {
  autoint n;
  vector<long long> a(n), b(n);
  cin >> a >> b;
  if (n == 1) answer(a[0] <= b[0] ? "YES\n" : "NO\n");
  if (test(a, b) == -1) answer("NO");
  long long x = test(a, add(0, -b[0], b));
  if (x != -1) answer(x + b[0] >= a[0] ? "YES" : "NO");
  long long l = 0, r = b[0];
  while (r - l > 1) {
    long long m = (l + r) / 2;
    (test(a, add(0, -m, b)) != -1 ? l : r) = m;
  }
  answer(l + test(a, add(0, -l, b)) >= a[0] ? "YES" : "NO");
}
