Codeforces Round #271 (Div. 2) d.flowers DP

D. Flowers

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 being tasty, there is a rule:marmot wants to eat white flowers only in groups of size K.

Now Marmot wonders on how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo 1000000007 (9 + 7).

Input

Input contains several test cases.

The first line contains integers t and k (1≤ t, k ≤105), where c9>T represents the number of test cases.

The next t lines contain II integers ai and b i (1≤ ai ≤ bi ≤105), describing the C30>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 ai and bi flowers at dinner modulo 1000000007 (9 + 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 F or example he can ' t eat (wwwr).

Test Instructions : There is a person like to eat flowers, there are white flowers, safflower two flowers, but eat white can only eat k flower or not to eat flowers, now ask you to eat N Flower program number is how much.

Solution: dp[i] means to eat I flower scheme number, then dp[i+k]+=dp[i] to update eat a white flower, dp[i+1]+=dp[i] used to update to eat a red flower

///1085422276#include <bits/stdc++.h>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#defineINF 0x7fffffffinline ll read () {ll x=0, f=1; CharCh=GetChar ();  while(ch<'0'|| Ch>'9')    {        if(ch=='-') f=-1; CH=GetChar (); }     while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; CH=GetChar (); }    returnx*F;}//****************************************#defineMAXN 100000+5#defineMoD 1000000007intDP[MAXN];intMain () {intN=read (), k=read ();     MEM (DP); dp[0]=1;  for(intI=0; i<=100000; i++){           if(i+1<=100000) Dp[i+1]= (dp[i+1]+dp[i])%MoD; if(i+k<=100000) Dp[i+k]= (Dp[i+k]+dp[i])%MoD;     } ll SUM[MAXN];     MEM (sum); sum[0]=0;  for(intI=1; i<=100000; i++) {Sum[i]+=sum[i-1]+Dp[i]; }     intb;  for(intI=1; i<=n;i++) {scanf ("%d%d",&a,&b); cout<< (sum[b]-sum[a-1])%mod<<Endl; }  return 0;}

Code

Codeforces Round #271 (Div. 2) d.flowers DP