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;
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long inf = LLONG_MAX / 10;
void ECHO(string _s) { cout << _s; }
void ECHO(bool _s) {
  if (_s)
    cout << "true";
  else
    cout << "false";
}
void ECHO(char _s) { cout << _s; }
void ECHO(long unsigned _s) { cout << _s; }
void ECHO(int _s) { cout << _s; }
void ECHO(long long _s) {
  if (_s == inf)
    cout << "inf";
  else
    cout << _s;
}
void ECHO(double _s) { cout << _s; }
void ECHO(long double _s) { cout << _s; }
void ECHO(bitset<8> b) { cout << b; }
template <typename T>
void ECHO(set<T> _s) {
  for (auto el : _s) {
    ECHO(el);
    cout << "_";
  }
}
template <typename T1, typename T2>
void ECHO(map<T1, T2> _s) {
  for (auto el : _s) {
    ECHO(el);
    cout << "_";
  }
}
template <typename T1, typename T2>
void ECHO(pair<T1, T2> _s) {
  cout << "(";
  ECHO(_s.first);
  cout << ",";
  ECHO(_s.second);
  cout << ")";
}
template <typename T>
void ECHO(vector<T> v) {
  for (auto el : v) {
    ECHO(el);
    cout << "_";
  }
}
template <typename T, typename... Args>
void ECHO(string _s, T x, Args... args) {
  int _i, c = 0;
  for (_i = 0; _i < (int)_s.size(); _i++) {
    if (c == 0 && _s[_i] == ',') break;
    if (_s[_i] == '(') c++;
    if (_s[_i] == ')') c--;
  }
  cout << " (" << _s.substr(0, _i) << "): ";
  ECHO(x);
  while (_i < (int)_s.size() && (_s[_i] == ',' || _s[_i] == ' ')) _i++;
  ECHO(_s.substr(_i), args...);
}
template <typename T>
class SegTree {
  vector<T> t, v;
  int n;
  T f(T a, T b) { return max(a, b); }
  void build(int k, int l, int r) {
    if (l == r) {
      t[k] = v[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
    t[k] = f(t[k << 1], t[k << 1 | 1]);
  }
  void update(int k, int l, int r, int p, T x) {
    if (l == r) {
      t[k] = x;
      return;
    }
    int mid = (l + r) >> 1;
    if (p <= mid)
      update(k << 1, l, mid, p, x);
    else
      update(k << 1 | 1, mid + 1, r, p, x);
    t[k] = f(t[k << 1], t[k << 1 | 1]);
  }
  T query(int k, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[k];
    if (qr < l || r < ql) return -inf;
    int mid = (l + r) >> 1;
    T qa = query(k << 1, l, mid, ql, qr);
    T qb = query(k << 1 | 1, mid + 1, r, ql, qr);
    if (qr <= mid) return qa;
    if (mid + 1 <= l) return qb;
    return f(qa, qb);
  }

 public:
  SegTree(vector<T> &_v) {
    v = _v;
    n = v.size();
    t.assign(4 * n, 0);
    build(1, 0, n - 1);
  }
  void update(int p, T x) { update(1, 0, n - 1, p, x); }
  T query(int ql, int qr) { return query(1, 0, n - 1, ql, qr); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long tt, n, i;
  cin >> tt;
  while (tt--) {
    cin >> n;
    vector<long long> a(n), b(n);
    long long sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    for (i = 0; i < n; i++) cin >> b[i], sum -= b[i];
    if (sum > 0) {
      cout << "NO\n";
      continue;
    }
    for (i = 0; i < n; i++) {
      a.push_back(a[i]);
      b.push_back(b[i]);
    }
    n *= 2;
    vector<long long> psum(n, 0);
    psum[0] = a[0] - b[0];
    for (i = 1; i < n; i++) psum[i] = psum[i - 1] + a[i] - b[i];
    SegTree<long long> st(psum);
    bool ok = true;
    if (st.query(0, n - 1) - b.back() > 0) ok = false;
    for (i = 1; i < n; i++) {
      if (st.query(i, i + n - 1) - psum[i - 1] - b[i - 1] > 0) ok = false;
    }
    if (ok)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
