Flowers

I - Flowers

Crawling in process... Crawling failed Time Limit:1500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 Input

Input

3 2
1 3
2 3
4 4

Output

6
5
5

Sample Output

      
      
      
      

       
       
       
        37 
      
      
      
      

 

Hint

• 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).

/*
 dp[i]=(dp[i-1]+dp[i-k])%MOD;
 正确的考虑情况：
    LL tmp=(sum[r]-sum[l-1])%MOD;
    if(tmp<0){
        tmp+=MOD;
        tmp%=MOD;
    }
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const int MOD= 1000000007 ;
const int maxn=100000+10;
LL  dp[maxn],sum[maxn];
LL  t,k,l,r;
int main(){
scanf("%I64d%I64d",&t,&k);
for(int i=0;i<k;i++){
    dp[i]=1;
}
for(int i=k;i<maxn;i++){
 dp[i]=(dp[i-1]+dp[i-k])%MOD;
}
for(int i=1;i<maxn;i++){
    sum[i]=(sum[i-1]+dp[i])%MOD;
}
while(t--){
    scanf("%I64d%I64d",&l,&r);
    LL tmp=(sum[r]-sum[l-1])%MOD;
    if(tmp<0){
        tmp+=MOD;
        tmp%=MOD;
    }
    printf("%I64d\n",tmp);
}
return 0;
}