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:
//@formatter:off
#include<bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < int(n); ++i)
#define rrep(i,n) for (int i = int(n)-1; i >= 0; i--)
#define rep2(i,s,n) for (int i = int(s); i < int(n); ++i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define pb push_back
#define eb emplace_back
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vd vector<double>
#define vvd vector<vector<double>>
#define vs vector<string>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vp vector<P>
#define vvp vector<vector<P>>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using LP = pair<ll,ll>;
template<class S,class T> istream& operator>>(istream &is,pair<S,T> &p) { return is >> p.first >> p.second; }
template<class S,class T> ostream& operator<<(ostream &os,const pair<S,T> &p) { return os<<'{'<<p.first<<","<<p.second<<'}'; }
template<class T> istream& operator>>(istream &is,vector<T> &v) { for(T &t:v){is>>t;} return is; }
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) { os<<'[';rep(i,v.size())os<<v[i]<<(i==int(v.size()-1)?"":","); return os<<']'; }
void Yes(bool b) { cout << (b ? "Yes" : "No") << '\n'; }
void YES(bool b) { cout << (b ? "YES" : "NO") << '\n'; }
template<class T> bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;
const ll linf = 1001001001001001001;
//@formatter:on

int to[30][200010];
ll sum[30][200010];

int rev_to[30][200010];
ll rev_sum[30][200010];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n, k;
    cin >> n >> k;
    vi x(n);
    cin >> x;
    rep(i, n) {
        auto it = lower_bound(all(x), x[i] + k);
        to[0][i] = it - x.begin();
        it = upper_bound(all(x), x[i] - k);
        rev_to[0][i] = it - x.begin() - 1;
        if (rev_to[0][i] == -1) rev_to[0][i] = n;
        sum[0][i] = rev_sum[0][i] = i;
    }
    to[0][n] = n;
    rev_to[0][n] = n;
    rep(i, 29) {
        rep(j, n+1) {
            to[i + 1][j] = to[i][to[i][j]];
            rev_to[i + 1][j] = rev_to[i][rev_to[i][j]];
            sum[i + 1][j] = sum[i][j] + sum[i][to[i][j]];
            rev_sum[i + 1][j] = rev_sum[i][j] + rev_sum[i][rev_to[i][j]];
        }
    }
    auto calc_to = [&](int st, int len) {
        rep(k, 30) {
            if (len >> k & 1) {
                st = to[k][st];
            }
        }
        return st;
    };
    auto calc_sum = [&](bool rev, int st, int len) {
        ll res = 0;
        rep(k, 30) {
            if (len >> k & 1) {
                if (rev) {
                    res += rev_sum[k][st];
                    st = rev_to[k][st];
                } else {
                    res += sum[k][st];
                    st = to[k][st];
                }
            }
        }
        return res;
    };
    int q;
    cin >> q;
    while (q--) {
        int l, r;
        cin >> l >> r;
        l--;
        r--;
        int ok = 0, ng = 1 << 20;
        auto f = [&](int x) -> bool {
            return calc_to(l, x) <= r;
        };
        while (abs(ok - ng) > 1) {
            int mid = (ng + ok) / 2;
            if (f(mid)) ok = mid;
            else ng = mid;
        }
        ll ans = calc_sum(true, r, ng) - calc_sum(false, l, ng) + ng;
        cout << ans << '\n';
    }
}