Mr. Ant has 3 boxes and the infinite number of marbles. Now he wants to know the number of ways he can put marbles in these three boxes when the following conditions hold.

1) Each box must contain at least 1 marble.

2) The summation of marbles of the 3 boxes must be in between X and Y inclusive.

Now you are given X and Y. You have to find the number of ways Mr. Ant can put marbles in the 3 boxes.

Input

Input starts with an integer T, denoting the number of test cases. Each test case contains two integers X and Y.

Constraints

1<=T<=1000000

1<=X<= Y<=1000000

Output

For each test case, print the required answer modulo 1000000007

Explanation for the first test case

Way 01

Way 02

Way 03

Way 04

Way 05

Way 06

Way 07

Way 08

Way 09

Note: use faster i/o method.

n個相同的東西放進三個不同的盒子裏，每個盒子至少要有一個

這就是裸的排列組合題

用隔板法很容易知道，對於一個單獨的n，答案就是C(n-1,2)，令f(x)=C(x-1,2)

對於f(x)的一段求和，顯然在n>=3時才有方案，即f(x)定義域x>=3

令g(x)=f(3)+f(4)+...+f(x)，那麽答案就是g(r)-g(l-1)，（考慮到定義域應當是g(r)-g(l)+f(l)不過似乎不這樣也行）

然後根據組合數的性質C(2,2)+C(3,2)+...+C(x-1,2)=C(x,3)

所以g(x)=C(x,3)

然後做完了

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 #define mod 1000000007
18 using namespace std;
19 inline LL read()
20 {
21     LL x=0,f=1;char ch=getchar();
22     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
23     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
24     return x*f;
25 }
26 inline LL calc(LL a)//return C a 3
27 {
28     if (a<3)return 0;
29     return a*(a-1)*(a-2)/6%mod;
30 }
31 int main()
32 {
33     int T=read();
34     while (T--)
35     {
36         LL a=read(),b=read();
37         printf("%lld\n",(calc(b)-calc(a-1)+mod)%mod);
38     }
39 }

Spoj-ANTP Mr. Ant & His Problem