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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
int sz(const T &a) {
  return (int)a.size();
}
const int MAXN = 1e6 + 1;
ll cost[MAXN];
ll cap[MAXN];
ll te[MAXN], te2[MAXN];
int n;
int check(ll am) {
  if (am > min(cost[0], cap[n - 1])) return 0;
  for (int i = 0; i < n; i++) {
    te[i] = cost[i], te2[i] = cap[i];
  }
  te[0] -= am;
  te2[n - 1] -= am;
  int ret = 0;
  for (int i = 0; i < n; i++) {
    te2[i] -= te[i];
    if (te2[i] < 0) return ret;
    if (i < n - 1) {
      ll x = min(te[i + 1], te2[i]);
      te[i + 1] -= x, te2[i] -= x;
      if (te2[i]) ret = -1;
    }
  }
  return 1;
}
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> cost[i];
    }
    for (int i = 0; i < n; i++) {
      cin >> cap[i];
    }
    ll lo = 0, hi = min(cost[0], cap[n - 1]);
    bool done = false;
    while (lo != hi) {
      ll mid = (lo + hi) / 2 + 1;
      int te = check(mid);
      if (te == 1) {
        printf("YES\n");
        done = true;
        break;
      } else if (te == -1)
        hi = mid - 1;
      else
        lo = mid;
    }
    if (!done) {
      if (check(lo) == 1 || check(lo + 1) == 1) {
        printf("YES\n");
        done = true;
      }
    }
    if (!done) {
      printf("NO\n");
    }
  }
  return 0;
}
