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;
bool __hack = std::ios::sync_with_stdio(false);
auto __hack1 = cin.tie(nullptr);
namespace template_util {
constexpr int bytecount(uint64_t x) { return x ? 1 + bytecount(x >> 8) : 0; }
template <int N>
struct bytetype {};
template <uint64_t N>
struct minimal_uint : bytetype<bytecount(N)> {};
}  // namespace template_util
template <class T>
T next(istream& in) {
  T ret;
  in >> ret;
  return ret;
}
template <class T>
vector<T> next_vec(istream& in, size_t n) {
  vector<T> ret(n);
  for (size_t i = 0; i < n; ++i) {
    ret[i] = next<T>(in);
  }
  return ret;
}
void solve(istream& in, ostream& out) {
  auto tests = next<int>(in);
  while (tests--) {
    auto n = next<int>(in);
    auto a = next_vec<int>(in, n);
    auto b = next_vec<int>(in, n);
    int le = 0, ri = b[0];
    bool ok = false;
    while (le <= ri) {
      int e = (le + ri) / 2;
      int nxt = max(0, a[1] - e);
      int needMore = b[0] - e;
      bool closeCircle = true;
      for (int i = (1); i < (n); ++i) {
        int fill = b[i] - nxt;
        if (fill < 0) {
          closeCircle = false;
          break;
        }
        nxt = max(0, a[(i + 1) % n] - fill);
      }
      if (closeCircle) {
        if (nxt <= needMore) {
          ok = true;
          break;
        }
        ri = e - 1;
      } else {
        le = e + 1;
      }
    }
    if (ok) {
      out << "YES\n";
    } else {
      out << "NO\n";
    }
  }
}
int main() {
  solve(cin, cout);
  return 0;
}
