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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define fi first
#define se second
#define debug printf("aaaaaaaaaaa\n");
const int maxn=1e6+5,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=0x3f3f3f3f3f3f3f3f;
ll n,l,r;
ll fac[maxn],finv[maxn];
ll qpow(ll a,ll b){
    ll ans=1,base=a;
    while(b){
        if(b&1) ans=ans*base%mod;
        base=base*base%mod;
        b=b>>1;
    }
    return ans;
}
void init(){
    fac[0]=finv[0]=1;
    for(int i=1;i<=2e5;i++){
        fac[i]=fac[i-1]*i%mod;
        finv[i]=qpow(fac[i],mod-2);
    }
}
ll c(int a,int b){
    ll ans=fac[a]*finv[b]%mod*finv[a-b]%mod;
    return ans;
}
int main(){
    init();
    int _;cin>>_;
    while(_--){
        cin>>n>>l>>r;
        int lim=min(r-n,1-l);
//        cout<<c(n,n/2)<<'\n';
        ll ans=c(n,n/2)*lim%mod;

        if(n%2==1) ans=2*ans%mod;
        for(int i=lim+1;;i++){
            int cnt1=max(1ll*0,n-(r-i+1)+1);
            int cnt2=max(1ll*0,l+i-1);
            if(n%2==0){
                if(cnt1>n/2||cnt2>n/2) break;
                ans=(ans+c(n-cnt1-cnt2,n/2-cnt1))%mod;
            }else{
                if(cnt1>n/2+1||cnt2>n/2+1) break;
                ans=(ans+c(n-cnt1-cnt2,n/2-cnt1))%mod;
                ans=(ans+c(n-cnt1-cnt2,n/2+1-cnt1))%mod;
            }
        }
        cout<<ans<<'\n';
    }
    return 0;
}
