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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
constexpr int inf=1e9+7;
constexpr ll longinf=1LL<<60 ;
constexpr ll mod=1e9+7 ;


int nxt[202020][20], pre[202020][20];
ll nsum[202020][20], psum[202020][20];
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n,k;
    cin>>n>>k;
    vector<int> a(n);
    rep(i,n)cin>>a[i];
    int j=0;
    rep(i,n){
        while(j < n && a[i]+k>a[j])j++;
        nxt[i][0]=j;
        nsum[i][0]=j;
    }
    j=n-1;
    for(int i=n-1;i>=0;--i){
        while(j>=0 && a[j]+k>a[i])j--;
        pre[i][0]=j;
        psum[i][0]=j;
    }
    rep(j,19)rep(i,n){
        if(nxt[i][j]!=n){
            nxt[i][j+1]=nxt[nxt[i][j]][j];
            nsum[i][j+1]=nsum[i][j]+nsum[nxt[i][j]][j];
        } else {
            nxt[i][j+1]=n;
        }
        if(pre[i][j]!=-1){
            pre[i][j+1]=pre[pre[i][j]][j];
            psum[i][j+1]=psum[i][j]+psum[pre[i][j]][j];
        } else {
            pre[i][j+1]=-1;
        }
    }
    int q;
    cin>>q;
    while(q--){
        int l,r;
        cin>>l>>r;
        --l;--r;
        ll lcnt = 1, rcnt = 1, lsum = l, cl = l, rsum = r, cr = r;
        for(int i=19;i>=0;i--){
            if(nxt[cl][i]<=r){
                lcnt += 1<<i;
                lsum += nsum[cl][i];
                cl = nxt[cl][i];
            }
        }
        for(int i=19;i>=0;i--){
            if(pre[cr][i]>=l){
                rcnt += 1<<i;
                rsum += psum[cr][i];
                cr = pre[cr][i];
            }
        }
        //assert(lcnt == rcnt);
        //cout << lcnt << " " <<lsum << " " << rsum <<endl;
        cout << rsum - lsum + lcnt << endl;
    }
    return 0;
}