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]. 
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#include <limits.h>
using namespace std;
 
#define mod 1000000007
#define ll long long
#define ld long double

#define fi first
#define se second

const int N = 2e5+10;

int f[N], inf[N];

int mul(int a, int b){
    return int(a*1ll * b % mod);
}

int binpow(int a, int k){
    int ans = 1;
    while(k>0){
        
        if(k & 1) ans = mul(ans, a);
        
        a = mul(a,a);
        k>>=1;
    }
    return ans;
}


void precal(){
    f[0] = 1;
    inf[0] = 1;
    
    for(int i= 1; i<N; i++){
        f[i] = mul(i, f[i-1]);
        inf[i] = binpow(f[i], mod-2);
    }
}


int C(int n, int k){
    if(k<0 || n<k) return 0;
    
    return mul(f[n], mul(inf[k], inf[n-k]));
}

inline void solve(){
    int n, l, r; cin>>n>>l>>r;
    
    int k = min(1-l, r-n);
    int half = n/2;
    int ans = mul(k, C(n, half));
    
    if(n & 1) ans = (ans + (ll)mul(k, C(n, half+1)))%mod;
    
    for(k++;;k++){
        int lf = max(1, l+k);
        int rg = min(n, r-k);
        
        if(rg-lf+1 < 0) break;
        
        ans = (ans + C(rg-lf+1, half-lf+1) )%mod;
        
        if(n&1) ans = (ans + C(rg-lf+1, half-lf+2) )%mod;
    }
    
    cout<<ans<<endl;
}
 
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    precal();
    
 
    int t; cin>>t;
    
    for(;t;t--)
    solve();
                        
}

