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 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.

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

• 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).

思路：

1、设定dp【i】表示长度为i的情况有多少合法放置方式。

dp【i】=dp【i-1】+dp【i-k】；

长度为i-1的时候，直接在其每个合法的放置方式的右边多加一个红色的花也都是合法的情况。

长度为i-k的时候，直接在其每个合法的放置方式的右边多加k个连续白色的花也都是合法的情况。

那么累加即可。

2、那么答案就是sum【bi】-sum【ai】

代码：

 #include<bits/stdc++.h>

 #define db double

 #include<vector>

 #define ll long long

 #define vec vector<ll>

 #define Mt  vector<vec>

 #define ci(x) scanf("%d",&x)

 #define cd(x) scanf("%lf",&x)

 #define cl(x) scanf("%lld",&x)

 #define pi(x) printf("%d\n",x)

 #define pd(x) printf("%f\n",x)

 #define pl(x) printf("%lld\n",x)

 const int N = 1e5 + ;

 const int mod = 1e9 + ;

 const int MOD = mod-;

 const db  eps = 1e-;

 const db  PI = acos(-1.0);

 using namespace std;

 ll f[N],sum[N];

 int main()

 {

     int n,k;

     ci(n),ci(k);

     memset(f,,sizeof(f));

     memset(sum,,sizeof(sum));

     f[]=;

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

         if(i>=k) f[i]=(f[i-]+f[i-k])%mod;

         else f[i]=f[i-]%mod;

         sum[i]=(sum[i-]+f[i])%mod;

     }

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

         int l,r;

         ci(l),ci(r);

         ll ans=(sum[r]-sum[l-]+mod)%mod;

         pl(ans);

     }

     return ;

 }