codeforces 474D

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 ai and bi (1 ≤ ai ≤ bi ≤ 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 ai and bi 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).

题目的意思是Marmot只吃连续的k个白色的花，红色的花任意吃，计算从a到b之间有多少种可能。

例如：

　　k=2时，长度为1Marmot只能吃1（R）

　　k=2时，长度为2Marmot只能吃2 （RR）和（WW）

　　k=2时，长度为3Marmot只能吃3（RRR）、（RWW）和（WWR）

l	1	2	3	4	5	6
sum	1	2	3	5	8	13

先枚举k=2的情况，然后发现数字之间的规律，dp[i]=dp[i-1]+dp[i-k]

然后题目要求大于1000000007的数要求余

 

#include<iostream>
#define mod 1000000007
#define MAXN 1000005
using namespace std;
typedef long long ll;
ll dp[MAXN];
ll sum[MAXN];
int main()
{
    int t, k, a, b;
    cin >> t >> k;
    for (int i = 0;i < k;i++)
    {
        dp[i] = 1;
        sum[i] = i;
    }
    sum[0] = 0;
    for (int i = k;i < MAXN;i++)
    {
        dp[i] = (dp[i - 1] + dp[i - k]) % mod;
        sum[i] = (sum[i - 1] + dp[i]) % mod;
    }
    for (int i = 0;i < t;i++)
    {
        cin >> a >> b;
        cout << (sum[b] - sum[a - 1] + mod) % mod << endl;
    }
    return 0;
}