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:
//#pragma GCC optimize("Ofast", "unroll-loops")
//#pragma GCC target("sse", "sse2", "sse3", "ssse3", "sse4", "avx")

#ifdef __APPLE__
# include <iostream>
# include <cmath>
# include <algorithm>
# include <stdio.h>
# include <cstdint>
# include <cstring>
# include <string>
# include <cstdlib>
# include <vector>
# include <bitset>
# include <map>
# include <queue>
# include <ctime>
# include <stack>
# include <set>
# include <list>
# include <random>
# include <deque>
# include <functional>
# include <iomanip>
# include <sstream>
# include <fstream>
# include <complex>
# include <numeric>
# include <immintrin.h>
# include <cassert>
# include <array>
# include <tuple>
# include <unordered_map>
# include <unordered_set>
# include <thread>
#else
# include <bits/stdc++.h>
#endif

#define F first
#define S second
#define MP make_pair
#define PB push_back

#define all(a) a.begin(),a.end()
#define len(a) (int)(a.size())
#define mp make_pair
#define pb push_back
#define fir first
#define sec second

using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;

const int max_n = 2e5+10, inf = 1000111222;
const int max_log=20;

int x[max_n];
pii prec_L[max_n][max_log];
pii prec_R[max_n][max_log];

int next_L[max_n];
int next_R[max_n];

int main() {
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n,k;
    cin>>n>>k;
    for (int i=0;i<n;i++){
        cin>>x[i];
    }

    auto precalc=[&]()
    {
        next_L[0]=0;
        for (int i=0;i<n;i++){
            if (i){
                next_L[i]=next_L[i-1];
            }
            while (next_L[i]<n && x[next_L[i]]-x[i]<k){
                next_L[i]++;
            }
//            cout<<"next_L["<<i<<"] :: "<<next_L[i]<<"\n";
        }

        next_R[n-1]=n-1;
        for (int i=n-1;i>=0;i--){
            if (i+1!=n){
                next_R[i]=next_R[i+1];
            }
            while (next_R[i]>=0 && x[i]-x[next_R[i]]<k){
                next_R[i]--;
            }
//            cout<<"next_R["<<i<<"] :: "<<next_R[i]<<"\n";
        }

        for (int j=0;j<max_log;j++){
            if (j==0){
                for (int i=0;i<n;i++){
                    prec_L[i][j]=mp(next_L[i],i);
                }
            }
            else{
                for (int i=0;i<n;i++){
                    int to = prec_L[i][j-1].fir;
                    if (to<n){
                        prec_L[i][j] = prec_L[to][j-1];
                        prec_L[i][j].sec += prec_L[i][j-1].sec;
                    }
                    else{
                        prec_L[i][j] = prec_L[i][j-1];
                    }
                }
            }
        }
        for (int j=0;j<max_log;j++){
            if (j==0){
                for (int i=0;i<n;i++){
                    prec_R[i][j]=mp(next_R[i],i);
                }
            }
            else{
                for (int i=0;i<n;i++){
                    int to = prec_R[i][j-1].fir;
                    if (to>=0){
                        prec_R[i][j] = prec_R[to][j-1];
                        prec_R[i][j].sec += prec_R[i][j-1].sec;
                    }
                    else{
                        prec_R[i][j] = prec_R[i][j-1];
                    }
                }
            }
        }
    };

    auto get_naive_ans_L=[&](int l,int r)
    {
        pii res=mp(1,0);
        int cur=l;
        for (int j=max_log-1;j>=0;j--){
            if (prec_L[cur][j].fir<=r){
                res.fir+=(1ll<<j);
                res.sec+=prec_L[cur][j].sec;
                cur=prec_L[cur][j].fir;
            }
        }
        res.sec+=cur;
        return res;
    };

    auto get_naive_ans_R=[&](int l,int r)
    {
        pii res=mp(1,0);
        int cur=r;
        for (int j=max_log-1;j>=0;j--){
            if (prec_R[cur][j].fir>=l){
                res.fir+=(1ll<<j);
                res.sec+=prec_R[cur][j].sec;
                cur=prec_R[cur][j].fir;
            }
        }
        res.sec+=cur;
        return res;
    };

    auto get_ans=[&](int l,int r)
    {
        pii ans_L=get_naive_ans_L(l,r);
        pii ans_R=get_naive_ans_R(l,r);
//        cerr<<l<<" "<<r<<" :: "<<"("<<ans_L.fir<<","<<ans_L.sec<<"), "<<"("<<ans_R.fir<<","<<ans_R.sec<<")"<<"\n";
        assert(ans_L.fir == ans_R.fir);

        return ans_L.fir + (ans_R.sec-ans_L.sec);
    };

    precalc();
    int q;
    cin>>q;
    while (q--){
        int l,r;
        cin>>l>>r;
        l--;
        r--;
        cout<<get_ans(l,r)<<"\n";
    }
}