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;
char buffer[2000020], *p1, *p2;
template <class T>
void read(T& x) {
  char ch = ((p1 == p2 &&
              (p2 = (p1 = buffer) + fread(buffer, 1, 2000020, stdin), p1 == p2))
                 ? EOF
                 : *p1++);
  x = 0;
  bool f = 1;
  while (!('0' <= ch && ch <= '9') && ch != '-')
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 2000020, stdin), p1 == p2))
              ? EOF
              : *p1++);
  if (ch == '-')
    f = 0,
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 2000020, stdin), p1 == p2))
              ? EOF
              : *p1++);
  while ('0' <= ch && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0',
    ch = ((p1 == p2 &&
           (p2 = (p1 = buffer) + fread(buffer, 1, 2000020, stdin), p1 == p2))
              ? EOF
              : *p1++);
  x = (f) ? x : -x;
}
int _num[20];
template <class T>
void write(T x) {
  if (!x) {
    putchar('0'), putchar('\n');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int c = 0;
  while (x) _num[++c] = x % 10, x /= 10;
  while (c) putchar('0' + _num[c--]);
  putchar('\n');
}
int n;
long long int a[2000020], b[2000020], c[2000020 << 1];
long long int st[2000020 << 1];
int head = 0, tail = 0;
inline void add(int i) {
  while (head > tail && c[i] >= c[st[head]]) head--;
  st[++head] = i;
}
inline void del(int i) {
  if (st[tail + 1] == i) tail++;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    head = tail = 0;
    scanf("%d", &n);
    long long int sa = 0, sb = 0;
    for (register int i = 1; i <= n; i++) scanf("%lld", a + i), sa += a[i];
    for (register int i = 1; i <= n; i++) scanf("%lld", b + i), sb += b[i];
    if (sa > sb) {
      printf("NO\n");
      continue;
    }
    for (register int i = 1; i <= n; i++) c[i] = a[i] - b[i], c[i + n] = c[i];
    for (register int i = 1; i <= 2 * n; i++) c[i] += c[i - 1];
    long long int mx = 0;
    for (register int i = 2; i <= n; i++) add(i);
    for (register int i = 1; i <= n; i++) {
      mx = max(mx, c[st[tail + 1]] - c[i] - b[i]);
      add(n + i), del(i + 1);
    }
    if (mx == 0)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
