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;
int gcd(int i, int j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 1e17;
const int Y = 2000100;
const long long m = 1000000007;
long long a[Y];
long long tree[4 * Y];
long long b[Y];
long long pl[Y];
long long f[Y];
int sz;
void add(long long v, int ps) {
  ps += sz;
  while (ps) {
    tree[ps] = min(tree[ps], v);
    ps /= 2;
  }
}
long long get(int ql, int qr, int v, int tl, int tr) {
  if (ql <= tl && tr <= qr) return tree[v];
  long long res = INF;
  int mid = (tr + tl) / 2;
  if (ql <= mid) res = min(res, get(ql, qr, 2 * v, tl, mid));
  if (qr > mid) res = min(res, get(ql, qr, 2 * v + 1, mid + 1, tr));
  return res;
}
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int t = getint<int>();
  while (t--) {
    int n = getint<int>();
    sz = 1;
    while (sz <= 2 * (n + 1)) sz <<= 1;
    for (int i = 0; i < 2 * sz; ++i) tree[i] = INF;
    for (int i = 0; i < n; ++i) {
      a[i] = getint<int>();
      a[i + n] = a[i];
    }
    for (int i = 0; i < n; ++i) {
      b[i] = getint<int>();
      b[i + n] = b[i];
    }
    f[0] = b[0] - a[0];
    add(f[0], 0);
    for (int i = 1; i < 2 * n; ++i) {
      f[i] = b[i] - a[i];
      long long ad = min(b[i - 1], a[i]);
      ad = min(ad, f[i - 1]);
      ad = max(ad, (long long)0);
      f[i] += ad;
      pl[i] = ad;
      add(f[i], i);
    }
    bool res = false;
    for (int i = 0; i < n && res == false; ++i) {
      if (pl[i] == 0) {
        res |= (get(i, i + n - 1, 1, 0, sz - 1) >= 0);
      }
    }
    if (res)
      cout << "YES\n";
    else {
      cout << "NO\n";
    }
  }
  return 0;
}
