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 long MOD = (long long)1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long cmb(long long r, long long c) {
  return fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
priority_queue<int, vector<int>, greater<int>> pq;
int n;
int a[1000000], b[1000000], tmp[1000000];
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &b[i]);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int l = 0, r = a[0], mid;
    while (l <= r) {
      mid = (l + r) / 2;
      int p = mid, k, flag = 0;
      for (int i = 1; i < n; i++) {
        k = p + a[i] - b[i];
        if (k < 0) {
          flag = 1;
          break;
        }
        p = min(k, a[i]);
      }
      if (flag)
        l = mid + 1;
      else
        r = mid - 1;
    }
    int p = l, k, flag = 0;
    if (p > a[0]) {
      printf("NO\n");
      continue;
    }
    for (int i = 1; i < n; i++) {
      k = p + a[i] - b[i];
      if (k < 0) {
        flag = 1;
        break;
      }
      p = min(k, a[i]);
    }
    if (p + (a[0] - l) >= b[0] && !flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
}
