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;
const long long int INF = 9223372036854775807;
const long long int mod = (long long int)1e9 + 7;
#pragma GCC target("avx2")
#pragma GCC optimization("unroll-loops")
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
struct point {
  long long int x;
  long long int y;
};
long long int cross(point a, point b) { return a.x * b.y - b.x * a.y; }
bool ccw(point p1, point p2) {
  if (cross(p1, p2) > 0) {
    return true;
  } else {
    return false;
  }
}
void ccw_sort(vector<point> &v) {
  vector<point> v1, v2;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].y > 0) {
      v1.push_back(v[i]);
    } else {
      v2.push_back(v[i]);
    }
  }
  sort(v1.begin(), v1.end(), ccw);
  sort(v2.begin(), v2.end(), ccw);
  for (int i = 0; i < v2.size(); i++) {
    v1.push_back(v2[i]);
  }
  v = v1;
}
bool check(int x, int y, int n, int m) {
  if (x >= 0 && y >= 0 & x < n && y < m) {
    return true;
  } else {
    return false;
  }
}
vector<vector<int> > adj;
vector<int> deg;
vector<int> w;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<long long int> a(n);
    long long int sum = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      sum = sum + a[i];
    }
    vector<long long int> b(n);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    vector<vector<long long int> > dp(n);
    for (int i = 0; i < n; i++) {
      dp[i].resize(2);
    }
    long long int ans = 1e18;
    dp[0][0] = 1e16;
    dp[0][1] = b[n - 1] + b[0] - a[0];
    for (int i = 1; i <= n - 1; i++) {
      if (i != n - 1) {
        dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = min(dp[i - 1][0] - a[i] + b[i] + b[i - 1],
                       dp[i - 1][1] - a[i] + b[i]);
      } else {
        dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = min(dp[i - 1][0] - a[i] + b[i - 1], dp[i - 1][1] - a[i]);
      }
    }
    ans = min(ans, sum + min(dp[n - 1][0], dp[n - 1][1]));
    dp[0][0] = 0;
    dp[0][1] = 1e16;
    for (int i = 1; i <= n - 1; i++) {
      if (i != n - 1) {
        dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = min(dp[i - 1][0] - a[i] + b[i] + b[i - 1],
                       dp[i - 1][1] - a[i] + b[i]);
      } else {
        dp[i][0] = min(dp[i - 1][0], dp[i - 1][1]);
        dp[i][1] = min(dp[i - 1][0] - a[i] + b[i - 1] + b[i],
                       dp[i - 1][1] - a[i] + b[i]);
      }
    }
    ans = min(ans, sum + min(dp[n - 1][0], dp[n - 1][1]));
    if (ans < sum) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
    }
  }
}
