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 n, t[5000005], save[5000005], a[1000005], b[1000005], pown, l, r, ad;
void solv(long long x) {
  for (int i = x - 1; i >= 1; i--) {
    a[i] -= b[i];
    a[i] = max(a[i], 0LL);
    long long ind = i - 1;
    if (!ind && x != 1) ind = n;
    if (a[i] > 0 && b[ind] < a[i]) {
      printf("NO\n");
      return;
    }
    b[ind] -= a[i];
  }
  for (int i = n; i >= x; i--) {
    a[i] -= b[i];
    a[i] = max(a[i], 0LL);
    long long ind = i - 1;
    if (i == x) ind = 0;
    if (a[i] > 0 && a[i] > b[ind]) {
      printf("NO\n");
      return;
    }
    b[ind] -= a[i];
  }
  printf("YES\n");
}
void add(long long x, long long L, long long R) {
  if (L > r || R < l) return;
  if (L >= l && R <= r) {
    t[x] += ad;
    save[x] += ad;
    return;
  }
  if (save[x]) {
    t[2 * x] += save[x];
    t[2 * x + 1] += save[x];
    save[2 * x] += save[x];
    save[2 * x + 1] += save[x];
    save[x] = 0;
  }
  add(2 * x, L, (L + R) / 2);
  add(2 * x + 1, (L + R) / 2 + 1, R);
  t[x] = t[2 * x];
  if (t[2 * x + 1] < t[x]) t[x] = t[2 * x + 1];
}
void solve() {
  scanf("%I64d", &n);
  pown = 1;
  while (pown <= n) pown *= 2;
  for (int i = 1; i <= 2 * pown; i++) {
    t[i] = 0;
    save[i] = 0;
  }
  long long asum = 0, bsum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    asum += a[i];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &b[i]);
    bsum += b[i];
  }
  for (int i = 1; i <= n; i++) {
    l = i, r = n, ad = b[i] - a[i];
    add(1, 1, pown);
  }
  if (t[1] >= 0) {
    solv(1);
    return;
  }
  for (int i = 1; i < n; i++) {
    l = 1, r = pown, ad = -(b[i] - a[i]);
    add(1, 1, pown);
    l = i, r = i, ad = -asum + a[i] + bsum - b[i] + (b[i] - a[i]);
    add(1, 1, pown);
    if (t[1] >= 0) {
      solv(i + 1);
      return;
    }
  }
  printf("NO\n");
}
int main() {
  long long q;
  cin >> q;
  while (q--) solve();
  return 0;
}
