D. Flowers Codeforces Round 271 (Div. 2)-优快云博客

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

Copy

output

Copy

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).

题目大致意思：放白花和红花，鼹鼠要至少有一个连续长度为k的W，问有多少方式。

经典dp题。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

const ll N = 1e5+7 ,M = 1e9+7;
ll n,m,k;
ll v[N];

void solve(){
	cin >> n >> m;
	cout << (v[m]-v[n-1]+M)%M << endl;
	return;
}

int main(){
	ll t=1;cin >> t >> k;
	v[0]=1;
	for(ll i = 1 ; i <= N ; i ++)
		v[i] = (v[i-1] + (i >= k ? v[i-k] : 0) )%M;
	for(ll i = 1 ; i <= N ; i ++)
		v[i] = (v[i] + v[i-1])%M;
	while(t--)solve(); 
	return 0;
}