#include <bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

typedef long long ll;

const int mod=1e9+;

const int N=1e5+;

/********组合数模板*********/

ll pow_mod(ll a, ll b) {

    ll res = 1ll; a %= mod;

    while(b){

        if(b & ) res = res * a % mod;

        a = a * a % mod;

        b >>= ;

    } return res;

}

ll inv(ll a) { return pow_mod(a, mod-); }

ll F[N], Finv[N];//F是阶乘，Finv是逆元的阶乘

void init() {

    F[] = Finv[] = 1ll;

    for(int i = ; i < N; i ++){

        F[i] = F[i-] * 1ll * (ll)i % mod;

        Finv[i] = Finv[i-] * 1LL * inv(i) % mod;

    }

} // O(n)预处理

ll C(ll n, ll m) {

    if(m <  || m > n) return ;

    return F[n] * 1LL * Finv[n - m] % mod * Finv[m] % mod;

} // O(1)获得组合数C(n,m)

/**************************/

int block[N];

ll res[N];

struct mo

{

    int n,m;

    int id,block;

    bool operator < (const mo &p) const{

      if(block==p.block) return n<p.n;

      else return block<p.block;

    }

}s[N];

int main()

{

    int T;

    scanf("%d",&T);

    int len=sqrt(N+0.5);

    for(int i=;i<T;i++)

    {

        scanf("%d %d",&s[i].n,&s[i].m);

        s[i].block=s[i].m/len;s[i].id=i;

    }

    sort(s,s+T);

    ll ans=;

    init();

    int L=,R=;

    for(int i=;i<T;i++)

    {

        int l=s[i].n,r=s[i].m;

        while(L>l) ans=((ans+C(L-1LL,R))%mod*Finv[])%mod, L--;

        while(L<l) ans=(*ans%mod-C(L,R)+mod)%mod, L++;

        while(R<r) ans=(ans+C(L,R+))%mod, R++;

        while(R>r) ans=(ans-C(L,R)+mod)%mod, R--;

        res[s[i].id]=ans;

    }

    for(int i=;i<T;i++)

        printf("%lld\n",res[i]);

    return ;

}