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;
}
constexpr long long MAX = 5100000;
constexpr long long INF = 1LL << 60;
constexpr int inf = 1000000007;
constexpr long long mod = 998244353LL;
const long double PI = acos((long double)(-1));
using namespace std;
struct network {
  long long min_l, min_r, max_r;
  network() : min_l(0), min_r(0), max_r(0) {}
  network(long long _min_l, long long _min_r, long long _max_r)
      : min_l(_min_l), min_r(_min_r), max_r(_max_r) {}
};
network get(long long a, long long b) {
  if (a <= b)
    return network(0, b - a, b);
  else
    return network(a - b, 0, b);
}
void merge(network& a, network b) {
  if (a.min_l == -1 or b.min_l == -1) {
    a = network(-1, -1, -1);
    return;
  }
  if (a.max_r < b.min_l) {
    a = network(-1, -1, -1);
    return;
  }
  if (a.min_r >= b.min_l)
    a = network(a.min_l, min(b.max_r, b.min_r + a.min_r - b.min_l),
                min(b.max_r, b.min_r - b.min_l + a.max_r));
  else
    a = network(a.min_l + b.min_l - a.min_r, b.min_r,
                min(b.max_r, b.min_r - b.min_l + a.max_r));
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<long long> a(n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  vector<long long> b(n);
  for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
  network nt = get(a[0], b[0]);
  for (int i = 1; i < n; i++) merge(nt, get(a[i], b[i]));
  if (nt.min_l == -1) {
    puts("NO");
    return;
  }
  if (nt.min_l <= nt.min_r)
    puts("YES");
  else
    puts("NO");
}
int main() {
  int kkt;
  scanf("%d", &kkt);
  while (kkt--) {
    solve();
  }
  return 0;
}
