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;

const int MAX_N = 2e5;
const int LMT = 1e9 + 7;

int n, l, r, ans;
array<int, MAX_N + 1> f;
array<int, MAX_N + 1> invf;

int powInt(int x, int p) {
    if(p == 0) return 1;
    int s = (p & 1) ? x : 1;
    int hf = powInt(x, p / 2);
    s = (long long) s * hf % LMT;
    s = (long long) s * hf % LMT;
    return s;
}

void precalc() {
    f[0] = 1;
    invf[0] = 1;
    for(int i = 1; i <= MAX_N; ++i) {
        f[i] = (long long) f[i - 1] * i % LMT;
        invf[i] = powInt(f[i], LMT - 2);
    }
}

void input() {
    cin >> n >> l >> r;
}

void process() {
    if(n & 1) {
        ans = (long long) min(1 - l, r - n) * f[n + 1] % LMT * invf[(n + 1) / 2] % LMT * invf[(n + 1) / 2] % LMT;
        for(int k = min(1 - l, r - n) + 1; true; ++k) {
            int lp = max(1, k + l);
            int rp = min(n, r - k);
            if(lp - 1 > (n + 1) / 2) break;
            if(n - rp > (n + 1) / 2) break;
            if(n - rp + lp - 1 > n) break;
            ans += (long long) f[rp - lp + 2] * invf[(n + 1) / 2 - lp + 1] % LMT * invf[rp - (n + 1) / 2 + 1] % LMT;
            ans %= LMT;
        }
        
    } else {
        ans = (long long) min(1 - l, r - n) * f[n] % LMT * invf[n / 2] % LMT * invf[n / 2] % LMT;
        for(int k = min(1 - l, r - n) + 1; true; ++k) {
            int lp = max(1, k + l);
            int rp = min(n, r - k);
            if(lp - 1 > n / 2) break;
            if(n - rp > n / 2) break;
            if(n - rp + lp - 1 > n) break;
            ans += (long long) f[rp - lp + 1] * invf[n / 2 - lp + 1] % LMT * invf[rp - n / 2] % LMT;
            ans %= LMT;
        }
    }
}

void output() {
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    // freopen("input.txt", "r", stdin);
    
    int t;
    
    precalc();
    for(cin >> t; t; --t) {
        input();
        process();
        output();
    }
    
    return 0;
}