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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2")
using namespace std;
using lld = long long int;
using pll = pair<lld, lld>;
using lf = double;
using pff = pair<lf, lf>;
using pii = pair<int, int>;
int n, m;
lld a[1000009];
lld b[1000009];
lld c[1000009];
lld d[1000009];
lld tmpa[1000009];
int main() {
  int i, j, k;
  int t = 1, tv = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      d[i] = 0;
    }
    for (i = 0; i < n; i++) {
      scanf("%lld", &b[i]);
    }
    lld tot = 0;
    lld minn = 0;
    for (i = 0; i < n; i++) {
      c[i] = b[i] - a[i];
      tot += c[i];
      if (minn > c[i]) minn = c[i];
    }
    if (tot < 0) {
      printf("NO\n");
      continue;
    }
    if (minn >= 0) {
      printf("YES\n");
      continue;
    }
    int mi = 0;
    for (i = 0; i < n; i++) {
      if (c[i] > 0) break;
    }
    mi = i;
    if (mi > 0) {
      for (i = 0; i < n; i++) {
        tmpa[i] = a[(i + mi) % n];
      }
      for (i = 0; i < n; i++) {
        a[i] = tmpa[i];
      }
      for (i = 0; i < n; i++) {
        tmpa[i] = b[(i + mi) % n];
      }
      for (i = 0; i < n; i++) {
        b[i] = tmpa[i];
      }
      for (i = 0; i < n; i++) {
        tmpa[i] = c[(i + mi) % n];
      }
      for (i = 0; i < n; i++) {
        c[i] = tmpa[i];
      }
    }
    mi = 0;
    bool valid = false;
    for (i = 0; i < n; i++) {
      c[i] = min(b[i], a[i]);
    }
    while (1) {
      valid = true;
      bool stable = true;
      for (i = 0; i < n; i++) {
        if (a[i] - c[i] - d[i] > 0) valid = false;
        if (a[i] - d[i] < b[i]) {
          lld diff = max(b[i] - max(a[i] - d[i], 0LL), 0LL);
          if (diff != d[(i + 1) % n]) stable = false;
          c[i] = min(b[i], max(a[i] - d[i], 0LL));
          d[(i + 1) % n] = diff;
        }
      }
      if (valid) break;
      if (stable) break;
    }
    printf("%s\n", valid ? "YES" : "NO");
  }
  return 0;
}
