【Codeforces Round 271 (Div 2)D】【DP】Flowers 黑色必须连续摆放k，长度为n的摆放方案数

在这款有趣的算法挑战中，Marmot需要按照特定规则享用红白花朵组成的晚餐。问题要求计算不同花朵数量下，符合规则的组合总数，并输出模10^9+7的结果。

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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 a_i and b_i (1 ≤ a_i ≤ b_i ≤ 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 between a_i and b_i 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).

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=1e5+10,M=0,Z=1e9+7,ms63=1061109567;
int f[N],s[N];
int n,k,l,r;
int main()
{
	while(~scanf("%d%d",&n,&k))
	{
		f[0]=s[0]=1;
		for(int i=1;i<=100000;++i)
		{
			f[i]=f[i-1];
			if(i>=k)f[i]=(f[i]+f[i-k])%Z;
			s[i]=(s[i-1]+f[i])%Z;
		}
		for(int i=1;i<=n;++i)
		{
			scanf("%d%d",&l,&r);
			printf("%d\n",(s[r]-s[l-1]+Z)%Z);
		}
	}
	return 0;
}
/*
【题意】
我们要用黑白棋子摆成一列，其中黑色棋子必须k个一组连续摆放。
问你，对于长度为[1,n]的摆放，最终有多少种方案。

【类型】
DP

【分析】
这题我们用f[i]表示长度为i的合法摆放方案有多少种。
然后，对于每次在位置i的摆放，可以考虑这个位置是摆放白棋还是黑棋。
对应的状态转移前驱分别是f[i-1]和f[i-k]
然后记录一个前缀和，我们这道题就做完啦！

【时间复杂度&&优化】
O（n）

*/