D. Flowers

time limit per test
1.5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of
them white and some of them red.

But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.

Now Marmot wonders in how many ways he can eat between a and b flowers.
As the number of ways could be very large, print it modulo 1000000007 (109 + 7).

Input

Input contains several test cases.

The first line contains two integers t and k (1 ≤ t, k ≤ 105),
where t represents the number of test cases.

The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105),
describing the i-th test.

Output

Print t lines to the standard output. The i-th
line should contain the number of ways in which Marmot can eat betweenai and bi flowers
at dinner modulo 1000000007 (109 + 7).

Sample test(s)

input

3 2
1 3
2 3
4 4

output

6
5
5

Note

For K = 2 and
length 1 Marmot can eat (R).

For K = 2 and
length 2 Marmot can eat (RR)
and (WW).

For K = 2 and
length 3 Marmot can eat (RRR),
(RWW) and (WWR).

For K = 2 and
length 4 Marmot can eat, for example, (WWWW)
or (RWWR), but for example he can't eat (WWWR).

解题说明：此题是一道动态规划题，由于要连续吃k朵白花，那么能够得到吃花的不同方式的转移方程，a[i]=a[i-1]+a[i-k]，要么是吃完前i-1朵花再吃一朵红花，要么是吃了i-k朵花再连续吃k朵白花。有了转移方程后就可以先打表，再根据输入的数据进行输出即可。

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int main()
{
int t,k,i;
int p,b;
long long int a[100001];a[0]=0;
long long int mod=1000000007;
cin>>t>>k;
for(i=1;i<=100000;i++)
{
if(i<k)
{
a[i]=1;
}
else if(i==k)
{
a[i]=2;
}
else
{
a[i]=(a[i-1]+a[i-k])%mod;
}
}
for(i=2;i<=100000;i++)
{
a[i]=(a[i-1]+a[i]);
}
while(t--)
{
cin>>p>>b;
cout<<(a[b]-a[p-1])%mod<<endl;
}
return 0;
}