$将m=R-L带入公式可得：C_{R-L+n+1}^{R-L+1}-1$
$数据范围1e9,使用lucas计算即可，复杂度为Plog_N^P(log可被忽略)$

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const int p = 1000003;

int qsm(int a, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (ll)res * a % p;
        a = (ll)a * a % p;
        k >>= 1;
    }
    return res;
}

int C(int a,int b){
    if(a<b) return 0;

    int down = 1, up = 1;  //up代表分子，down代表分母
    for (int i = a, j = 1; j <= b; i --, j ++ )
    {
        up = (ll)up * i % p;
        down = (ll)down * j % p;
    }

    return (ll)up * qsm(down, p - 2) % p;
}

int Lucas(int a,int b){
    if(a<p&&b<p) return C(a,b)%p;
    else return (ll)C(a%p,b%p)*Lucas(a/p,b/p)%p; 
}

int main()
{
    int T;
    cin >> T;
    while (T -- )
    {
        int n, l, r;
        cin >> n >> l >> r;
        cout << (Lucas(r - l + n + 1, r - l + 1) - 1 + p ) % p << endl;
    }

    return 0;
}