#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7,I=1e9+10;
int n,k,a[201000];
#define pi pair<int,int>
#define fi first
#define se second
struct SGT{
    pi mi[801000];
    void sc(int p){mi[p]=min(mi[p<<1],mi[p<<1|1]);}
    inline void build(int p,int l,int r,int ty){
        if(l==r){mi[p]={a[l]*ty,l};return;}
        int mid=(l+r)>>1;
        build(p<<1,l,mid,ty),build(p<<1|1,mid+1,r,ty);
        sc(p);
    }
    inline void up(int p,int l,int r,int x,int z){
        if(l==r){mi[p]={z,l};return;}
        int mid=(l+r)>>1;
        if(x<=mid)up(p<<1,l,mid,x,z);
        else up(p<<1|1,mid+1,r,x,z);
        sc(p);
    }
    inline pi ask(int p,int l,int r,int x,int y){
        if(x<=l&&r<=y)return mi[p];
        int mid=(l+r)>>1;
        pi nt={I,-1};
        if(x<=mid)nt=min(nt,ask(p<<1,l,mid,x,y));
        if(y>mid)nt=min(nt,ask(p<<1|1,mid+1,r,x,y));
        return nt;
    }
}A,B;
struct BIT{
    int tr[201000];
    inline void ad(int x,int z){for(;x<=n;x+=(x&-x))tr[x]+=z;}
    inline int ask(int x){int s=0;for(;x;x-=(x&-x))s+=tr[x];return s;}
}T;
void cl(int x){A.up(1,1,n,x,I),B.up(1,1,n,x,-I),T.ad(x,-1);}
int ans,jc[200100],ij[200100],inv[201000];
#define pb push_back
int C(int a,int b){if(a<0||b<0||a<b)return 0;return 1ll*jc[a]*ij[b]%mod*ij[a-b]%mod;}
void cdq(int l,int r){
    if(l>r)return;
    pi t1=A.ask(1,1,n,l,r);if(t1.fi>=I)return;
    pi t2=B.ask(1,1,n,l,r);
    int wc=T.ask(r)-T.ask(l-1);
    if(t1.fi-t2.fi<=k){
        //找到所有最小值并删去。
        int o=l;
        vector<int>P;
        while(o<=r){
            pi z=A.ask(1,1,n,o,r);
            if(z.fi==t1.fi)P.pb(z.se),o=z.se+1;
            else break;
        }
        for(int x:P)cl(x);
        ans=1ll*ans*C(wc,(int)P.size())%mod;
        cdq(l,r);return;
    }
    else{
        int o=l;
        vector<int>P;
        while(o<=r){
            pi z=B.ask(1,1,n,o,r);
            if(z.fi==t2.fi)P.pb(z.se),o=z.se+1;
            else break;
        }
        int sz=P.size();
        for(int i=0;i<=sz;i++)cdq((i==0)?l:(P[i-1]+1),(i==sz)?r:(P[i]-1));
    }
}
int main(){
    scanf("%d%d",&n,&k);
    ans=1;
    inv[1]=1;
    for(int i=2;i<=n;i++)inv[i]=mod-1ll*inv[mod%i]*(mod/i)%mod;
    jc[0]=ij[0]=1;
    for(int i=1;i<=n;i++)jc[i]=1ll*i*jc[i-1]%mod,ij[i]=1ll*inv[i]*ij[i-1]%mod,T.ad(i,1);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    A.build(1,1,n,1),B.build(1,1,n,-1);
    cdq(1,n);printf("%d",ans);
    return 0;
}