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 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).

又是一道求前缀和的dp，dp[i]=dp[i]+dp[i-k];

给你一组范围a---b，求你只能连续吃k朵白花，红花随便，问有几种吃法

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#define sf(a) scanf("%d",&a)
#define rep(i,a,b) for(i=a;i<=b;i++)
#define e 1e-8
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int idata=1e5+5;
const ll mod=1e9+7;

int i,j,k;
int judge,flag;
//vector<int> step(idata);
ll sum[idata];
ll dp[idata];
ll n,m,t;
int maxx=0,minn=inf;
ll cnt,len,ans;
priority_queue< int, vector<int> , greater<int> >q;

int main()
{
    while(cin>>t>>n)
    {
        dp[0]=1;
        //一开始没初始化，还以为方程求错了
        for(i=1;i<=1e5+5;i++)
        {
            if(n>i)
            {
                dp[i]=1;
                sum[i]=sum[i-1]+dp[i];
            }
            else
            {
                dp[i]=(dp[i-1]+dp[i-n])%mod;
                sum[i]=(sum[i-1]+dp[i])%mod;
            }
        }
        while(t--)
        {
            int a,b;
            cin>>a>>b;
            cout<<(mod+sum[b]-sum[a-1])%mod<<endl;
            //多加一个mod的原因：sum[a-1]和sum[b]在mod过程中，可能出现sum[a-1]大的情况
            //为什么是a-1，最后一组样列提示了一下，要求第4组，就要用前4组－前3组
        }
    }
    return 0;
}