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.
import sys

input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp():
    return (int(input()))
def instr():
    return (str(input()))
def inlt():
    return (list(map(int, input().split())))
def insr():
    s = input()
    return(list(map(int, list(s[:len(s) - 1]))))
# def insr():
#     s = input()
#     return list(s[:len(s) - 1])
def invr():
    return (map(int, input().split()))


from collections import Counter

def check(cities, stations, k, allow):
    # cities = cities.copy()
    n = len(cities)
    k = min(cities[0], k)
    last_st = stations[-1] - k
    c_i = cities[0] - k
    for i in range(n - 1):
        d = stations[i] - (c_i + cities[i + 1])
        if d > 0:
            # cities[i + 1] = 0
            c_i = 0
            allow -= d
            if allow < 0:
                return 1
        elif stations[i] < c_i:
            return -1
        else:
            c_i = cities[i + 1] - (stations[i] - c_i)
    if c_i > last_st:
        return -1
    return 0


def bin_search(cities, stations, allow, l, r):
    while l <= r:
        mid = l + (r - l) // 2
        res = check(cities, stations, mid, allow)
        if res == 0:
            return mid
        elif res == -1:
            l = mid + 1
        else:
            r = mid - 1
    return -1


def main():
    t = inp()
    for _ in range(t):
        n = inp()
        cities = inlt()
        stations = inlt()
        allow = sum(stations) - sum(cities)
        if allow < 0:
            print('NO')
        else:
            res = bin_search(cities, stations, allow, 0, stations[-1] + 1)
            if res == -1:
                print('NO')
            else:
                print('YES')


if __name__ == '__main__':
    main()
