Problem B. Harvest of Apples

Problem Description

There are n apples on a tree, numbered from 1 to n.
Count the number of ways to pick at most m apples.

Input

The first line of the input contains an integer T (1≤T≤105) denoting the number of test cases.
Each test case consists of one line with two integers n,m (1≤m≤n≤105).

Output

For each test case, print an integer representing the number of ways modulo 109+7.

Sample Input

2

5 2

1000 500

Sample Output

16

924129523

#include
using namespace std;


//现在比较水，没理解到根，先不写题解了
typedef long long ll;
#define rep(i,a,b) for(int i=a;i=a;i--)
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
const int maxn=1e5+10;

const int mod=1e9+7;

ll block;
struct Query{
    ll i,l,r;
    bool operator<(const Query& obj)const{
        if(l/block==obj.l/block)return r>=1;
    }
    return ans%mod;
}

//!!!inv的线性求法
void init(){
    fac[0]=1;
    for(int i=1;i=0;i--){
        inv[i]=inv[i+1]*(i+1)%mod;
    }
}

ll C(ll n,ll m){
    if(n==m||m==0)return 1;
    if(m>n)return 0;
    //printf("fac(%lld):%lld inv(%lld):%lld inv(%lld):%lld\n",r,fac[r],l,inv[l],r-l,inv[r-l]);
    return (((fac[n]*inv[m])%mod)*inv[n-m])%mod;
}
/*
1
3 1
3 2
*/
int main(){
    init();
   //printf("C(%d,%d):%lld\n",3,4,C(3,4));
    int m;
    scanf("%d",&m);
    ll max_dis=0;
    for(int i=0;iquery[i].l){
            l--;
            res=((res+C(l,r))%mod*inv[2])%mod;
        }
        while(rquery[i].r){
            res=(res-C(l,r)+mod)%mod;
            r--;
        }
        ans[query[i].i]=res;
    }
    for(int i=0;i