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;
long long gcd(long long aa, long long bb) {
  if (bb == 0)
    return aa;
  else
    return gcd(bb, aa % bb);
}
long long lcm(long long a, long long b) {
  if (a * b == 0) return 0;
  return (a * b) / gcd(a, b);
}
long long prime(long long x) {
  if (x <= 1) return 0;
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long check_in(long long x, long long y, long long n, long long m) {
  return x >= 0 && x < n && y >= 0 && y < m;
}
long long dx[4] = {1, -1, 0, 0};
long long dy[4] = {0, 0, 1, -1};
long long n, m, w;
const long long MAXN = 2005;
const long long MAXM = 1000005;
long long a[MAXM], b[MAXM];
long long ta[MAXM], tb[MAXM];
long long check(long long x) {
  long long i;
  for (i = 1; i <= n; i++) {
    ta[i] = a[i];
    tb[i] = b[i];
  }
  tb[1] -= x;
  ta[1] -= x;
  if (tb[1] < 0) {
    return 0;
  }
  for (i = 2; i <= n; i++) {
    long long mi = min(ta[i], tb[i - 1]);
    ta[i] -= mi;
    tb[i - 1] -= mi;
    tb[i] -= ta[i];
    if (tb[i] < 0) {
      return 0;
    }
  }
  tb[n] -= ta[1];
  ta[1] = 0;
  if (tb[n] < 0) {
    return 2;
  }
  return 1;
}
void solve() {
  cin >> n;
  long long i;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
  long long l = 0, r = a[1];
  while (l <= r) {
    long long mid = (l + r) / 2;
    long long x = check(mid);
    if (x == 1) {
      cout << "YES"
           << "\n";
      return;
    }
    if (x == 0) {
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << "NO"
       << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}
