Let's call an integer array a_1, a_2, ..., a_n good if a_i ≠ i for each i.

Let F(a) be the number of pairs (i, j) (1 ≤ i < j ≤ n) such that a_i + a_j = i + j.

Let's say that an array a_1, a_2, ..., a_n is excellent if: 

  * a is good; 
  * l ≤ a_i ≤ r for each i; 
  * F(a) is the maximum possible among all good arrays of size n. 



Given n, l and r, calculate the number of excellent arrays modulo 10^9 + 7.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. 

The first and only line of each test case contains three integers n, l, and r (2 ≤ n ≤ 2 ⋅ 10^5; -10^9 ≤ l ≤ 1; n ≤ r ≤ 10^9).

It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the number of excellent arrays modulo 10^9 + 7.

Example

Input


4
3 0 3
4 -3 5
42 -33 55
69 -42 146


Output


4
10
143922563
698570404

Note

In the first test case, it can be proven that the maximum F(a) among all good arrays a is equal to 2. The excellent arrays are: 

  1. [2, 1, 2]; 
  2. [0, 3, 2]; 
  3. [2, 3, 2]; 
  4. [3, 0, 1]. 
//Praise our lord and saviour qlf9
//DecimalFormat f = new DecimalFormat("##.00");

import java.util.*;
import java.io.*;
import java.math.*;
import java.text.*;
public class D{
public static long mod = 1000000007;
public static void main(String[] omkar) throws Exception
{
   BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
   StringTokenizer st = new StringTokenizer(in.readLine());
   StringBuilder sb = new StringBuilder();
   int cases = Integer.parseInt(st.nextToken());
   for(int i = 0; i < cases; i++)
   {
      solve(in, st, sb);
   }  
   System.out.println(sb);
   
   
        }
 public static void solve(BufferedReader in, StringTokenizer st, StringBuilder sb) throws Exception
 {
  st = new StringTokenizer(in.readLine());
   int n  = Integer.parseInt(st.nextToken());
   int l  = Integer.parseInt(st.nextToken());
   int r  = Integer.parseInt(st.nextToken());
   long[] facts = new long[n+1];
   long[] factsInv = new long[n+1];
   facts[0] = 1L;
   for(int i = 1; i <= n; i++)
   {
      facts[i] = (facts[i-1]*(long)(i))%mod;
      factsInv[i] = (long)(exp((int)(facts[i]), (int)(mod-2), (int)(mod)));
   } 
   factsInv[0] = factsInv[1];
   long tot = facts[n];
   tot = (tot*factsInv[n/2])%mod;
   tot = (tot*factsInv[n-n/2])%mod;
   long val = Math.min(Math.abs(l-1), Math.abs(r-n));
   long diff = (long)(Math.abs(Math.abs(l-1)-Math.abs(r-n)));
   tot = (tot*val)%mod;
   long x = (long)(n/2);
   long y = (long)(n-n/2);
   long w, z, temp;
   if(n%2 == 1)
   {
      tot = (tot*2L)%mod;
   }
   for(int i = 1; i<=n; i++)
   {
      w = (long)(x-i);
      z = y-(long)(Math.max(i-diff, 0));
      if(w>= 0L && z>= 0L)
      {
         temp = facts[(int)(w+z)];
         temp = (temp*factsInv[(int)(w)])%mod;
         temp = (temp*factsInv[(int)(z)])%mod;
         tot = (tot+temp)%mod;
      }
      if(n%2 == 1)
      {
         w = (long)(y-i);
      z = x-(long)(Math.max(i-diff, 0));
      if(w>= 0L && z>= 0L)
      {
         temp = facts[(int)(w+z)];
         temp = (temp*factsInv[(int)(w)])%mod;
         temp = (temp*factsInv[(int)(z)])%mod;
         tot = (tot+temp)%mod;
      }
      }
   }
   sb.append(tot+"\n");
 }
 public long fact(int a, int mod)
 {
   long c = (long)(1);
   for(int i = 1; i <= a; i++)
   {
      c = (c*(long)(i))%((long)(mod));
   }  
   return c;
 }
 public static int[] readArr(int N, BufferedReader in, StringTokenizer st) throws Exception
      {
         int[] arr = new int[N];
         st = new StringTokenizer(in.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         return arr;
      }
 public static int exp(int a, int exp, int mod)
{
   int e = exp;
   if(e==0)
   {
      return 1;
   }
   long b = (long)(a);
   long ret = 1;
   while(e > 1)
   {
      if(e % 2 == 0)
      {
         e = e/2;
         b = (b*b)%mod;
      }
      else
      {
         ret = (ret*b)%mod;
         e = e/2;
         b = (b*b)%mod;
      }
   }
   b = (b*ret)%mod;
   return (int)(b);
}


}