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>
# define LL long long
# define N 200010
# define mod 1000000007LL
using namespace std;

int T,n,l,r;
LL jc[N],inv[N],ans;

void add (LL& a,LL b) {a += b; a >= mod? a-= mod:0;}

LL powe(int x,LL y){
    LL ret = 1;
    while (y) {
        if (y&1LL) ret = ret*x%mod;
        x = 1LL*x*x%mod;
        y >>= 1;
    }
    // printf("%lld ",ret);
    return ret;
}

void pre() {
    jc[0] = 1;
    inv[0] = 1;
    for (int i=1; i<N; ++i) 
        jc[i] = jc[i-1]*i%mod,
        inv[i] = powe(jc[i],mod-2);
}

LL C(int n,int m) {
    if (m < 0 || n < m) return 0;
    return jc[n]*inv[m]%mod*inv[n-m]%mod;
}

int main() {
    scanf("%d",&T);
    pre();
    while (T--) {
        scanf("%d%d%d",&n,&l,&r);
        LL st = min(1-l,r-n);
        if (n&1) ans = ((C(n,n/2)+C(n,n/2+1)) % mod * st) % mod;
        else ans = C(n,n/2) * st % mod;
        int ll=0,rr=n+1;
        for (LL i=st+1; ; ++i) {
            if (ll + 1 - i < l) ll++;
            if (rr - 1 + i > r) rr--;
            if (ll >= rr) break;
            if (n/2-ll+1 < 0) break;
            if (n&1) {
                add (ans,C(rr - ll - 1,n/2-ll));
                add (ans,C(rr - ll - 1,n/2-ll+1));
            }
            else {
                add (ans,C(rr - ll - 1,n/2-ll));
            }
        }
        printf("%lld\n",ans);
    }
}