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 andb_i (1 ≤ a_i ≤ b_i ≤ 10⁵), describing the i-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

655

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).

考虑第n个。假如n是小于k的，那么仅仅能都是是R，也就是仅仅有一种情况。

假如大于等于k。假设第n个是W，那么从n-k+1到n所有为W，假设第n个是R，那么数量就是前n-1个的数量。

dp[n] = 1; (0<= n < k)

dp[n] = dp[n-1] + dp[n-k]; (n >= k)

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            #define mem(f) memset(f,0,sizeof(f))#define M 100005#define mod 1000000007#define MAX 0X7FFFFFFF#define maxn 100005#define lson o<<1, l, m#define rson o<<1|1, m+1, rusing namespace std;typedef long long LL;int n = maxn, k, t, a, b, dp[maxn], sum[maxn];int main(){ scanf("%d%d", &t, &k); for(int i = 0; i < k; i++) dp[i] = 1; for(int i = k; i < n; i++) dp[i] = (dp[i-1] + dp[i-k])%mod; for(int i = 1; i < n; i++) sum[i] = (sum[i-1] + dp[i])%mod; while(t--) { scanf("%d%d", &a, &b); printf("%d\n", ((sum[b]-sum[a-1])%mod+mod)%mod ); } return 0;}