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.
line = input()
t = int(line)
for _ in range(t):
    line = input()
    n = int(line)
    line = input()
    need = [int(i) for i in line.split(' ')]
    line = input()
    station =[int(i) for i in line.split(' ')]
    if sum(station) < sum(need):
        print('NO')
        continue
    lo, hi = 0, station[-1]
    flag = True
    xmax, cur, x = station[-1], 0, station[-1]
    for i in range(n - 1):
        cur = cur + station[i] - need[i]
        if cur < 0:
            xmax += cur
            x += cur
            cur = 0
        cur = min(cur, station[i])
        xmax = min(xmax, station[i] - cur)
        if xmax < 0:
            flag = False
            break
    # xmax, cur = station[1], 0
    # for i in range(1, n ):
    #     cur = cur + station[i] - need[i]
    #     if cur < 0:
    #         xmax += cur
    #         cur = 0
    #     cur = min(cur, station[i])
    #     xmax = min(xmax, station[i] - cur)
    #     if xmax < 0:
    #         flag = False
    #         break
    # while lo <= hi:
    #     mid = lo + (hi - lo) // 2
    #     cur = mid
    #     flag = True
    #     for i in range(n - 1):
    #         cur = cur + station[i] - need[i]
    #         if cur < 0:
    #             lo = mid + 1
    #             flag = False
    #             break
    #         cur = min(cur, station[i])
    #     if flag:
    #         if station[-1] - mid + cur >= need[-1]:
    #             break
    #         else:
    #             flag = False
    #             hi = mid - 1
    if flag and x + cur >= need[-1]:
        print('YES')
    else:
        print('NO')

            
            

