#include <bits/stdc++.h>
using namespace std;
const long long sz = 1e6 + 10;
long long a[sz], cap[sz], n;
long long solve(long long bck) {
  bool house = 1;
  long long sum = bck, last = cap[n] - bck;
  for (long long i = 1; i <= n;) {
    if (house) {
      sum -= a[i];
      sum = min(sum, 0LL);
    } else {
      sum += cap[i];
      if (sum < 0) return -1;
      i++;
    }
    if (house && i == n) break;
    house ^= 1;
  }
  sum += last;
  if (sum < 0) return -2;
  return 1;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    for (long long i = 1; i <= n; ++i) scanf("%lld", &cap[i]);
    long long lo = 0, hi = min(a[1], cap[n]), ok = 0;
    while (lo <= hi) {
      long long mid = lo + hi >> 1;
      long long got = solve(mid);
      if (got == -1)
        lo = mid + 1;
      else if (got == -2)
        hi = mid - 1;
      else {
        ok = 1;
        break;
      }
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long sz = 1e6 + 10;
long long a[sz], cap[sz], n;
long long solve(long long bck) {
  bool house = 1;
  long long sum = bck, last = cap[n] - bck;
  for (long long i = 1; i <= n;) {
    if (house) {
      sum -= a[i];
      sum = min(sum, 0LL);
    } else {
      sum += cap[i];
      if (sum < 0) return -1;
      i++;
    }
    if (house && i == n) break;
    house ^= 1;
  }
  sum += last;
  if (sum < 0) return -2;
  return 1;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
    for (long long i = 1; i <= n; ++i) scanf("%lld", &cap[i]);
    long long lo = 0, hi = min(a[1], cap[n]), ok = 0;
    while (lo <= hi) {
      long long mid = lo + hi >> 1;
      long long got = solve(mid);
      if (got == -1)
        lo = mid + 1;
      else if (got == -2)
        hi = mid - 1;
      else {
        ok = 1;
        break;
      }
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```