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;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
//const int MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const double pi = acos(-1.0);
const double EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

int main(){
    int N, K;
    cin >> N >> K;
    int X[N];
    rep(i, N) cin >> X[i];
    int nl[N][20], nr[N][20];
    ll sl[N][20], sr[N][20];
    int ptr = 0;
    rep(i, N){
        while(ptr < N && X[ptr] < X[i]+K) ptr++;
        nl[i][0] = ptr, sl[i][0] = i;
    }
    ptr = N-1;
    rep3(i, N-1, 0){
        while(ptr >= 0 && X[ptr] > X[i]-K) ptr--;
        nr[i][0] = ptr, sr[i][0] = i+1;
    }
    rep2(j, 1, 19){
        rep(i, N){
            if(nl[i][j-1] == N){
                nl[i][j] = N;
                sl[i][j] = sl[i][j-1];
            }
            else{
                nl[i][j] = nl[nl[i][j-1]][j-1];
                sl[i][j] = sl[i][j-1]+sl[nl[i][j-1]][j-1];
            }
        }
        rep(i, N){
            if(nr[i][j-1] == -1){
                nr[i][j] = -1;
                sr[i][j] = sr[i][j-1];
            }
            else{
                nr[i][j] = nr[nr[i][j-1]][j-1];
                sr[i][j] = sr[i][j-1]+sr[nr[i][j-1]][j-1];
            }
        }
    }
    int Q;
    cin >> Q;
    while(Q--){
        int l, r; cin >> l >> r; l--, r--;
        ll suml = 0, sumr = 0;
        int pl = l, pr = r;
        rep3(i, 19, 0){
            if(nl[pl][i] <= r){
                suml += sl[pl][i], pl = nl[pl][i];
            }
        }
        suml += pl;
        rep3(i, 19, 0){
            if(nr[pr][i] >= l){
                sumr += sr[pr][i], pr = nr[pr][i];
            }
        }
        sumr += pr+1;
        cout << sumr-suml << endl;
    }
}