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
using namespace std;
#define mod 1000000007
int t;
ll f[200005], n, l, r, k, ans;
ll ksm(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1)ans = ans * a % mod;
        a = a * a % mod, b >>= 1;
    }
    return ans;
}
ll C(ll m, ll n)
{
    return m < n || m < 0 || n < 0 ? 0 : f[m] * ksm(f[n], mod - 2) % mod * ksm(f[m - n], mod - 2) % mod;
}
int main()
{
    scanf("%d", &t); f[0] = 1;
    for (int i = 1; i <= 200000; ++i)f[i] = f[i - 1] * i % mod;
    while (t--)
    {
        scanf("%lld%lld%lld", &n, &l, &r);
        k = min(1 - l, r - n), ans = k * C(n, n / 2) * (n & 1 ? 2 : 1) % mod;
        for (++k;; ++k)
        {
            ll lf = max(1ll, l + k), rg = min(n, r - k);           
            if (lf > rg + 1)break;
            ans = (ans + C(rg - lf + 1, n / 2 - lf + 1) + (n & 1 ? C(rg - lf + 1, n / 2 - lf + 2) : 0)) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

