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>
ostream& operator<<(ostream& os, vector<T> V) {
  for (auto v : V) os << v << " ";
  return cout << "";
}
template <class T>
ostream& operator<<(ostream& os, set<T> S) {
  for (auto s : S) os << s << " ";
  return cout << "";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << P.first << " " << P.second;
}
const long long mod = 1e9 + 7;
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pwr(long long a, long long b) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int a[1000010], b[1000010], n;
int check(int x) {
  int i;
  int rem = b[1] - x;
  for (i = 2; i <= n; i++) {
    if (rem + b[i] < a[i]) return 0;
    rem = b[i] - max(a[i] - rem, 0);
  }
  if (rem + x < a[1]) return 2;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t, i;
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= n; i++) cin >> b[i];
    int l = 0, r = min(a[1], b[1]);
    bool z = 0;
    while (true) {
      if (l > r) break;
      int mid = (l + r) / 2;
      int val = check(mid);
      if (val != 1 and l == r) break;
      if (val == 1) {
        z = 1;
        break;
      }
      if (val == 2) {
        l = mid + 1;
      } else if (val == 0) {
        r = mid - 1;
      }
    }
    cout << (z ? "YES" : "NO") << "\n";
  }
}
