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>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

#define p_ary(ary,a,b) do { cout << "["; for (int count = (a);count < (b);++count) cout << ary[count] << ((b)-1 == count ? "" : ", "); cout << "]\n"; } while(0)
#define p_map(map,it) do {cout << "{";for (auto (it) = map.begin();;++(it)) {if ((it) == map.end()) {cout << "}\n";break;}else cout << "" << (it)->first << "=>" << (it)->second << ", ";}}while(0)

template<typename T1,typename T2>ostream& operator<<(ostream& os,const pair<T1,T2>& a) {os << "(" << a.first << ", " << a.second << ")";return os;}

const char newl = '\n';

int main() {
    int n,k,q;
    cin >> n >> k;
    vector<int> x(n);
    for (int i = 0;i < n;++i) cin >> x[i];
    vector<int> edge1(n),edge2(n,-1);
    vector<vector<pair<int,ll>>> anc1(n),anc2(n);
    for (int i = 0;i < n;++i) edge1[i] = lower_bound(x.begin(),x.end(),x[i]+k)-x.begin();
    for (int i = 0;i < n;++i) if (edge1[i] < n) edge2[edge1[i]] = i;
    for (int i = 1;i < n;++i) edge2[i] = max(edge2[i],edge2[i-1]);
    for (int i = 0;i < n;++i) if (edge1[i] < n) anc1[i].push_back({edge1[i],edge1[i]});
    for (int i = 0;i < n;++i) if (edge2[i] >= 0) anc2[i].push_back({edge2[i],edge2[i]+1});
    for (int i = 0;i < 20;++i) {
        for (int j = 0;j < n;++j) {
            if (anc1[j].size() < i+1 || anc1[anc1[j][i].first].size() < i+1) continue;
            anc1[j].push_back(anc1[anc1[j][i].first][i]);
            anc1[j].back().second += anc1[j][i].second;
        }
    }
    for (int i = 0;i < 20;++i) {
        for (int j = 0;j < n;++j) {
            if (anc2[j].size() < i+1 || anc2[anc2[j][i].first].size() < i+1) continue;
            anc2[j].push_back(anc2[anc2[j][i].first][i]);
            anc2[j].back().second += anc2[j][i].second;
        }
    }
    cin >> q;
    while (q--) {
        int l,r;
        cin >> l >> r;
        l--;r--;
        ll ans = r+1-l;
        int i = l,j = anc1[l].size();
        while (j) {
            j--;
            if (anc1[i].size() < j+1 || anc1[i][j].first > r) continue;
            ans -= anc1[i][j].second;
            i = anc1[i][j].first;
        }
        i = r,j = anc2[r].size();
        while (j) {
            j--;
            if (anc2[i].size() < j+1 || anc2[i][j].first < l) continue;
            ans += anc2[i][j].second;
            i = anc2[i][j].first;
        }
        cout << ans << newl;
    }
}