Problem: There are N points on a number line, i-th of which is placed on coordinate X_i. These points are numbered in the increasing order of coordinates. In other words, for all i (1 \leq i \leq N-1), X_i < X_{i+1} holds. In addition to that, an integer K is given.

Process Q queries.

In the i-th query, two integers L_i and R_i are given. Here, a set s of points is said to be a good set if it satisfies all of the following conditions. Note that the definition of good sets varies over queries.

* Each point in s is one of X_{L_i},X_{L_i+1},\ldots,X_{R_i}.
* For any two distinct points in s, the distance between them is greater than or equal to K.
* The size of s is maximum among all sets that satisfy the aforementioned conditions.



For each query, find the size of the union of all good sets.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^9
* 0 \leq X_1 < X_2 < \cdots < X_N \leq 10^9
* 1 \leq Q \leq 2 \times 10^5
* 1 \leq L_i \leq R_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
X_1 X_2 \cdots X_N
Q
L_1 R_1
L_2 R_2
\vdots
L_Q R_Q


Output

For each query, print the size of the union of all good sets in a line.

Examples

Input

5 3
1 2 4 7 8
2
1 5
1 2


Output

4
2


Input

15 220492538
4452279 12864090 23146757 31318558 133073771 141315707 263239555 350278176 401243954 418305779 450172439 560311491 625900495 626194585 891960194
5
6 14
1 8
1 13
7 12
4 12


Output

4
6
11
2
3

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define fi first
#define se second
#define lc (x<<1)
#define rc (x<<1|1)
#define gc getchar()//(p1==p2&&(p2=(p1=buf)+fread(buf,1,size,stdin),p1==p2)?EOF:*p1++)
#define mk make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define IT iterator 
#define vi vector<int>
#define TP template<class o>
#define SZ(a) ((int)a.size())
#define all(a) a.begin(),a.end()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N=2e5+10,size=1<<20,mod=998244353,inf=2e9;

//char buf[size],*p1=buf,*p2=buf;
template<class o> void qr(o &x) {
    char c=gc; x=0; int f=1;
    while(!isdigit(c)){if(c=='-')f=-1; c=gc;}
    while(isdigit(c)) x=x*10+c-'0',c=gc;
    x*=f;
}
template<class o> void qw(o x) {
    if(x/10) qw(x/10);
    putchar(x%10+'0');
}
template<class o> void pr1(o x) {
    if(x<0)x=-x,putchar('-');
    qw(x); putchar(' ');
}
template<class o> void pr2(o x) {
    if(x<0)x=-x,putchar('-');
    qw(x); putchar('\n');
}

int n,m,q,lg,x[N],r[N][20],l[N][20];
ll sl[N][20],sr[N][20];

int main() {
    qr(n); qr(m);
    for(int i=1;i<=n;i++) qr(x[i]);
    for(int i=1;i<=n;i++) {
        r[i][0]=r[i-1][0];
        while(r[i][0]<=n&&x[r[i][0]]-x[i]<m) r[i][0]++;
        sr[i][0]=r[i][0]-1;
    }
    l[n+1][0]=n;
    for(int i=n;i;i--) {
        l[i][0]=l[i+1][0];
        while(l[i][0]&&x[i]-x[l[i][0]]<m) l[i][0]--;
        sl[i][0]=l[i][0];
    }
    for(int i=0;i<20;i++) r[n+1][i]=n+1;
    for(int j=1;(1<<j)<=n;j++,lg++)
        for(int i=1;i<=n;i++) {
            l[i][j]=l[l[i][j-1]][j-1];
            sl[i][j]=sl[i][j-1]+sl[l[i][j-1]][j-1];
            r[i][j]=r[r[i][j-1]][j-1];
            sr[i][j]=sr[i][j-1]+sr[r[i][j-1]][j-1];
        }
    qr(q); while(q--) {
        int L,R; qr(L); qr(R); ll ans=0;
        ans -= L-1;
        for(int v=L,i=lg;i>=0;i--)
            if(r[v][i]<=R) 
				ans -= sr[v][i],v=r[v][i];
        ans += R; 
        for(int v=R,i=lg;i>=0;i--)
            if(l[v][i]>=L) 
				ans += sl[v][i],v=l[v][i];
        pr2(ans);
    }
    return 0;
}