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 int long long
#define fi first
#define se second

using namespace std;
const int MOD = 1e9 + 7;
const int INF = 1e16;
int fact[200005], inv[200005];
int n, L, R;

int POW(int a, int b)
{
    if (b == 0) return 1;
    int T = POW(a, b >> 1);
    T = (T * T) % MOD;
    return (b & 1 ? (T * a) % MOD : T);
}

void prep()
{
    fact[0] = inv[0] = 1;
    for (int i = 1; i <= 200000; ++i)
    {
        fact[i] = (fact[i-1] * i) % MOD;
        inv[i] = POW(fact[i], MOD - 2);
    }
}

int C(int n, int k)
{
    if (k > n || k < 0) return 0;
    return (fact[n] * ((inv[k] * inv[n-k]) % MOD)) % MOD;
}

void xuly()
{
    int kall = min(1 - L, R - n);
    int res = (C(n, n/2) * kall) % MOD;
    if (n & 1) res = (res + C(n, n/2 + 1) * kall) % MOD;
    for (int k = kall + 1;; ++k)
    {
        int l = max(1LL,L+k);
        int r = min(n,R-k);
        if (r - l + 1 < 0) break;
        int c = n / 2 - l + 1;
        res = (res + C(r - l + 1, c)) % MOD;
        if (n & 1) res = (res + C(r - l + 1, c + 1)) % MOD;
    }
    cout << res << '\n';
}

void nhap()
{
    cin >> n >> L >> R;
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    prep();
    int T; cin >> T;
    while (T--)
    {
        nhap();
        xuly();
    }

    return 0;
}
