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(2)
#pragma GCC optimize(3)
using namespace std;
inline int read() {
  int x = 0, neg = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') neg = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = 10 * x + op - '0';
    op = getchar();
  }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 2000005;
const long long inf = -1e18;
struct Edge {
  int to, nxt, val;
} edge[N];
int head[N], tot;
void add(int u, int v, int w) {
  edge[++tot] = {v, head[u], w};
  head[u] = tot;
}
int a[N], b[N], n;
int L(int x) { return x > 1 ? x - 1 : n; }
int R(int x) { return x < n ? x + 1 : 1; }
long long dis[N];
int vis[N];
void spfa(int St) {
  queue<int> q;
  for (register int i = 1; i <= n + 1; i++) vis[i] = 0, dis[i] = inf;
  q.push(St), dis[St] = 0, vis[St] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 0;
    for (register int i = head[u]; i; i = edge[i].nxt) {
      int v = edge[i].to;
      if (dis[v] < dis[u] + edge[i].val) {
        dis[v] = dis[u] + edge[i].val;
        if (!vis[v]) {
          vis[v] = 1;
          q.push(v);
        }
      }
    }
  }
}
void solve() {
  n = read();
  long long tott = 0;
  for (register int i = 1; i <= n; i++) a[i] = read(), tott += a[i];
  for (register int i = 1; i <= n; i++) b[i] = read(), tott -= b[i];
  if (tott > 0) {
    puts("NO");
    return;
  }
  for (register int i = 1; i <= n + 1; i++) head[i] = 0;
  tot = 0;
  for (register int i = 1; i <= n; i++) add(L(i), i, a[i] - b[L(i)]);
  for (register int i = 1; i <= n; i++) add(n + 1, i, 0);
  spfa(n + 1);
  for (register int i = 1; i <= n; i++) {
    if (dis[i] < 0 || dis[i] > b[i]) {
      puts("NO");
      return;
    }
  }
  puts("YES");
}
int main() {
  int T = read();
  while (T--) solve();
  return 0;
}
