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>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<P, ll> T;
typedef pair<long double , ll> Ps;
typedef pair<ll, bool> Pb;
typedef pair<ll, vector<ll>> Pd;

const ll INF = 3e18;
const ll fact_table = 3200008;
long double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll, vector<ll>, greater<ll> > pqls;
priority_queue <P, vector<P>, greater<P> > pqps;
//small priority queue
//top pop

ll dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
ll dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
//↓,→,↑,←


#define endl "\n"
#ifdef ENJAPMA
#undef endl
#endif

#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)();
#define pc(x) cout << x << ",";
#define rep(i, n) for (ll i = 0; i < (n); i ++)

typedef vector<ll> vec;
typedef vector<vector<ll>> mat;

const ll mod = 998244353ll;
// const ll mod = 1000000007ll;

ll mypow(ll a, ll b, ll m = mod) {ll x = 1; while (b) {while (!(b & 1)) {(a *= a) %= m; b >>= 1;}(x *= a) %= m; b--;} return x;}
vec readvec(ll read) { vec res(read); for (int i = 0; i < read; i++) { cin >> res[i]; } return res;}
void YES(bool cond) { if (cond) { p("YES");} else { p("NO");} return;}
void Yes(bool cond) { if (cond) { p("Yes");} else { p("No");} return;}
void line() { p("--------------------"); return;}

/*
ll fact[fact_table + 5], rfact[fact_table + 5];

void c3_init() {
    fact[0] = rfact[0] = 1;
    for (ll i = 1; i <= fact_table; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table], mod - 2, mod);
    for (ll i = fact_table; i >= 1; i--) {
        rfact[i - 1] = rfact[i] * i;
        rfact[i - 1] %= mod;
    }
    return;
}
ll c3(ll n, ll r) {
    return (((fact[n] * rfact[r]) % mod ) * rfact[n - r]) % mod;
}
*/

struct Timer {
    int64_t start; const int64_t CYCLES_PER_SEC = 2800000000;
    Timer() { reset(); }
    void reset() { start = getCycle(); }
    inline double get() { return (double)(getCycle() - start) / CYCLES_PER_SEC; }
    inline int64_t getCycle() {
        uint32_t low, high; __asm__ volatile ("rdtsc" : "=a" (low), "=d" (high));
        return ((int64_t)low) | ((int64_t)high << 32);
    }
};

bool multicase = false;

ll n, k;
ll x[200005], toright[200005][65], toleft[200005][65];
ll torightvalue[200005][65], toleftvalue[200005][65];

void precalc() {
    set<P> st;
    for (int i = 1; i <= n; i++) {
        st.insert(P(x[i], i));
    }

    // 番兵
    st.insert(P(INF, INF));
    st.insert(P(-INF, -INF));

    for (int i = 1; i <= n; i++) {
        ll right = (*st.lower_bound(P(x[i] + k, -INF))).second;
        toright[i][0] = right;
        torightvalue[i][0] = (right != INF) ? i : i;

        auto ite = st.lower_bound(P(x[i] - k + 1, -INF));
        ite --;
        ll left = (*ite).second;
        toleft[i][0] = left;
        toleftvalue[i][0] = (left != -INF) ? i + 1 : i + 1;
    }




    for (int i = 0; i < 60; i++) {
        for (int j = 1; j <= n; j++) {

            if (toright[j][i] == INF) {
                toright[j][i + 1] = INF;
            } else {
                toright[j][i + 1] = toright[toright[j][i]][i];
                torightvalue[j][i + 1] = torightvalue[j][i] + torightvalue[toright[j][i]][i];
            }

            if (toleft[j][i] == -INF) {
                toleft[j][i + 1] = -INF;
            } else {
                toleft[j][i + 1] = toleft[toleft[j][i]][i];
                toleftvalue[j][i + 1] = toleftvalue[j][i] + toleftvalue[toleft[j][i]][i];
            }
        }
    }
    /*
    for (int j = 1; j <= 20; j++) {
        p(j);
        for (int i = 1; i <= n; i++) {
            pe(toleft[i][j]);
            p(toright[i][j]);
        }
    }


    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= 20; j++) {
            pe(toleft[i][j]);
            p(toright[i][j]);
        }
    }
    */

}

void solve() {

    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
        cin >> x[i];
    }
    precalc();
    ll q;
    cin >> q;
    for (int i = 0; i < q; i++) {
        ll l, r;
        cin >> l >> r;
        ll il = l, ir = r;
        ll haba = 59;
        ll ans = 0;
        ll sum1 = 0, sum2 = 0;
        while (true) {
            // haba進んでも大丈夫なら進む
            if (toleft[r][haba] >= il && toright[l][haba] <= ir) {
                ans += toleftvalue[r][haba] - torightvalue[l][haba];
                sum1 += toleftvalue[r][haba];
                sum2 += torightvalue[l][haba];
                l = toright[l][haba];
                r = toleft[r][haba];
            } else {
                if (haba == 0) {
                    break;
                } else {
                    haba --;
                }
            }
        }
        /*
        p(ans);
        pe(l);p(r);
        pe(toleftvalue[r][0]);
        p(torightvalue[l][0]);
        */
        ans += toleftvalue[r][0] - torightvalue[l][0];
        p(ans);
    }

    return;
}

int main() {
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    ll q, testcase = 1;
    if (multicase) {
        cin >> q;
    } else {
        q = 1;
    }
    while (q--) {
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }

    // solve();
    return 0;
}