#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int a[N], b[N], c[N];
int n;
int over(int x) {
  if (x == n + 1) return 1;
  return x;
}
int f(int x) {
  for (int i = 1; i <= n; i++) c[i] = a[i];
  c[1] -= x;
  c[2] -= b[1] - x;
  for (int i = 2; i <= n; i++) {
    if (c[i] > 0) {
      if (b[i] >= c[i]) {
        c[over(i + 1)] -= b[i] - c[i];
        c[i] = 0;
      } else
        return -1;
    } else
      c[over(i + 1)] -= b[i];
  }
  if (c[1] <= 0) return 1;
  return -2;
}
inline void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++) read(b[i]);
    int l = 0, r = min(a[1], b[1]);
    bool flag = false;
    while (l < r) {
      int mid = (l + r) >> 1;
      int tmp = f(mid);
      if (tmp > 0) {
        flag = true;
        break;
      } else if (tmp == -1)
        r = mid - 1;
      else
        l = mid + 1;
    }
    if (f(l) > 0) flag = true;
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int N = 1000005;
int a[N], b[N], c[N];
int n;
int over(int x) {
  if (x == n + 1) return 1;
  return x;
}
int f(int x) {
  for (int i = 1; i <= n; i++) c[i] = a[i];
  c[1] -= x;
  c[2] -= b[1] - x;
  for (int i = 2; i <= n; i++) {
    if (c[i] > 0) {
      if (b[i] >= c[i]) {
        c[over(i + 1)] -= b[i] - c[i];
        c[i] = 0;
      } else
        return -1;
    } else
      c[over(i + 1)] -= b[i];
  }
  if (c[1] <= 0) return 1;
  return -2;
}
inline void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++) read(b[i]);
    int l = 0, r = min(a[1], b[1]);
    bool flag = false;
    while (l < r) {
      int mid = (l + r) >> 1;
      int tmp = f(mid);
      if (tmp > 0) {
        flag = true;
        break;
      } else if (tmp == -1)
        r = mid - 1;
      else
        l = mid + 1;
    }
    if (f(l) > 0) flag = true;
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

```