Codeforces Round #271 (Div. 2) D 简单dp

D. Flowers

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Examples

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).

题意：放置'R' 'W'   W必须连续放置t个才合法  问你放置[a,b]个 共有多少种方案  具体看样例

题解：dp[i] 表示长度为i的合法方案数目  转移方程 dp[i]=dp[i-1]+dp[i-t]

#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define ll __int64
#define mod 1000000007
using namespace std;
int k,t;
ll sum[100005];
ll dp[100005];
int main()
{
    scanf("%d %d",&k,&t);
    sum[0]=0;
    dp[0]=1;
    for(int i=1;i<t;i++){
        dp[i]=1;
        sum[i]=(sum[i-1]+dp[i])%mod;
    }
    for(int i=t;i<=100000;i++){
        dp[i]=(dp[i-1]+dp[i-t])%mod;
        sum[i]=(sum[i-1]+dp[i])%mod;
    }
    int a,b;
    for(int i=1;i<=k;i++){
        scanf("%d %d",&a,&b);
        printf("%I64d
",(sum[b]-sum[a-1]+mod)%mod);
    }
    return 0;
}