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 mp make_pair
#define pb push_back
#define x first
#define y second
typedef pair<int,int> pii;
typedef long long ll;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x,T y){x<y?x=y:T();}
template <typename T> void chkmin(T &x,T y){x>y?x=y:T();}
template <typename T> void readint(T &x)
{
	x=0;int f=1;char c;
	for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=x*10+c-'0';
	x*=f;
}
const int MOD=998244353;
inline int dmy(int x){return x>=MOD?x-MOD:x;}
inline void inc(int &x,int y){x=dmy(x+y);}
int qmi(int x,int y)
{
	int ans=1;
	for(;y;y>>=1,x=1ll*x*x%MOD)
		if(y&1)ans=1ll*ans*x%MOD;
	return ans;
}
const int MAXN=200005,MAXK=19;

int n,K,a[MAXN],l[MAXN][MAXK],r[MAXN][MAXK];
ll sl[MAXN][MAXK],sr[MAXN][MAXK];

int main()
{
	#ifdef LOCAL
	freopen("code.in","r",stdin);
//	freopen("code.out","w",stdout);
	#endif
	int Q,u,v;
	readint(n),readint(K);
	for(int i=1;i<=n;++i)readint(a[i]);
	a[0]=a[1]-K,a[n+1]=a[n]+K;
	for(int i=1;i<=n;++i)
		l[i][0]=upper_bound(a,a+n+2,a[i]-K)-a-1,
		sl[i][0]=(l[i][0]<=n?l[i][0]:0),
		r[i][0]=lower_bound(a,a+n+2,a[i]+K)-a,
		sr[i][0]=(r[i][0]<=n?r[i][0]:0);
	for(int j=0;j<MAXK;++j)r[n+1][j]=n+1;
	for(int j=1;j<MAXK;++j)
		for(int i=1;i<=n;++i)
			l[i][j]=l[l[i][j-1]][j-1],r[i][j]=r[r[i][j-1]][j-1],
			sl[i][j]=sl[i][j-1]+sl[l[i][j-1]][j-1],sr[i][j]=sr[i][j-1]+sr[r[i][j-1]][j-1];
	readint(Q);
	while(Q--)
	{
		readint(u),readint(v);
		ll su=u,sv=v;
		for(int i=u,j=MAXK-1;j>=0;--j)
			if(r[i][j]<=v)su+=sr[i][j],i=r[i][j];
		for(int i=v,j=MAXK-1;j>=0;--j)
			if(l[i][j]>=u)sv+=sl[i][j]+(1<<j),i=l[i][j];
		printf("%lld\n",sv-su+1);
	}
	return 0;
}