CF#271 (Div. 2) D.（dp）

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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 sizek.

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 modulo1000000007 (10⁹ + 7).

Input

Input contains several test cases.

The first line contains two integers t andk (1 ≤ t, k ≤ 10⁵), wheret 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 thei-th test.

Output

Print t lines to the standard output. Thei-th line should contain the number of ways in which Marmot can eat betweena_i andb_i flowers at dinner modulo1000000007 (10⁹ + 7).

Sample test(s)

Input

3 21 32 34 4

Output

Note

For K = 2 and length1 Marmot can eat (R).
For K = 2 and length2 Marmot can eat (RR) and (WW).
For K = 2 and length3 Marmot can eat (RRR), (RWW) and (WWR).
For K = 2 and length4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).

解题思路：

dp求解。题意是说给一个k值，这个人有两种选择，要么吃连续k个白花，要么不吃白花，改吃红花。问这个人在a~b这个区间内，吃的情况有多少种组合。

最开始的思路就是按每个区间的排列组合打表，然后在a~b区间累加查询。后来规律找错了······WA了两发·······不难发现，当k = 1时，它是按2 ^ i 分布的；当k != 1时，最开始推出k > i 时， dp[i] = 1，k < i 时，斐波那契分布··············最后发现这是错的········赛后大牛给的dp方程：dp[ i ] = dp[ i - 1] + dp[ i - k ]。

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;__int64 dp[100002];__int64 sum[100002];int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    int n, k;    while(~scanf("%d%d",&n,&k))    {        memset(dp , 0 , sizeof(dp));        memset(sum , 0 , sizeof(sum));        for(int i = 0 ; i <= 100001 ; i ++)        {            if(i < k)            {                dp[i] = 1;                sum[i] = sum[i-1] + dp[i];            }            else            {                dp[i] = dp[i-1] + dp[i-k];                dp[i] %= MOD;                sum[i] = sum[i-1] + dp[i];                sum[i] %= MOD;            }        }        int a , b;        for(int i = 0 ; i < n ; i ++)        {            scanf("%d%d",&a,&b);            printf("%I64d\n", (sum[b] - sum[a-1] + MOD) % MOD);        }    }}

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