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 java.util.*;
import java.io.*;

public class NetworkCoverage {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        while(t-->0){
            int n = Integer.parseInt(f.readLine());
            long[] cities = new long[n];
            long[] stations = new long[n];
            StringTokenizer st1 = new StringTokenizer(f.readLine());
            StringTokenizer st2 = new StringTokenizer(f.readLine());
            for(int i = 0; i < n; i++){
                cities[i] = Long.parseLong(st1.nextToken());
                stations[i] = Long.parseLong(st2.nextToken());
            }
            boolean pos = false;
            long lo = -1L;
            long hi = stations[0]+1;
            while(hi > lo+1){
                long mid = (lo+hi)/2;
                int resp = possible(n, cities, stations, mid);
                if(resp == 0){
                    pos = true;
                    break;
                }
                if(resp == -1){
                    lo = mid;
                }else{
                    hi = mid;
                }
            }
            if(pos) out.println("YES");
            else out.println("NO");

        }
        out.close();
    }
    public static int possible(int n, long[] cities, long[] stations, long val1){
        long prevStat = stations[0]-val1;
        for(int i = 1; i < n; i++){
            if(cities[i] > prevStat + stations[i]) return 1;
            prevStat = Math.min(stations[i], prevStat+stations[i]-cities[i]);
        }
        if(prevStat < cities[0]-val1) return -1;
        return 0;
    }
}
