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;
mt19937 gene(233);
inline char GET_CHAR() {
  const int maxn = 131072;
  static char buf[maxn], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxn, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline long long fastpo(long long x, long long n, long long mod) {
  long long res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
inline string itoa(int x, int width = 0) {
  string res;
  if (x == 0) res.push_back('0');
  while (x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  while ((int)res.size() < width) res.push_back('0');
  reverse(res.begin(), res.end());
  return res;
}
const int _B = 131072;
char buf[_B];
int _bl = 0;
inline void flush() {
  fwrite(buf, 1, _bl, stdout);
  _bl = 0;
}
__inline void _putchar(char c) {
  if (_bl == _B) flush();
  buf[_bl++] = c;
}
inline void print(long long x, char c) {
  static char tmp[20];
  int l = 0;
  if (!x)
    tmp[l++] = '0';
  else {
    while (x) {
      tmp[l++] = x % 10 + '0';
      x /= 10;
    }
  }
  for (int i = l - 1; i >= 0; i--) _putchar(tmp[i]);
  _putchar(c);
}
const int N = 1000033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int a[N], b[N];
int main() {
  int t;
  t = getInt();
  for (int qq = 1; qq <= t; qq++) {
    int n;
    n = getInt();
    for (int i = 1; i <= n; i++) {
      a[i] = getInt();
    }
    for (int i = 1; i <= n; i++) {
      b[i] = getInt();
    }
    int le = -1, ri = min(a[1], b[1]);
    bool ans = false;
    while (le != ri) {
      int mid = (le + ri + 1) / 2;
      int rmn = b[1] - mid;
      bool brk = false;
      for (int i = 2; i <= n; i++) {
        if (rmn + b[i] < a[i]) {
          brk = 1;
          break;
        } else {
          rmn = b[i] - max(0, (a[i] - rmn));
        }
      }
      if (brk) {
        ri = mid - 1;
      } else {
        le = mid;
        if (mid + rmn >= a[1]) {
          ans = true;
        }
      }
    }
    printf("%s\n", ans ? "YES" : "NO");
  }
}
