#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define pii pair<int,int>
#define f first
#define s second
#define chmin(a, b) a = min(a,b)
#define chmax(a, b) a = max(a,b)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define all(x) x.begin(),x.end()
#define v vector
ll mod = 1e9+7;
ll mul(ll a, ll b) {return a*b%mod;}
void mul2(ll& a, ll b) {a = mul(a,b);}
ll exp(ll a, ll b) {
ll res = 1;
while (b) {
if (b&1) {res = mul(res,a);}
a = mul(a,a);
b /= 2;
} return res;
}
const int MAX_N = 2e5;
int a[MAX_N], small_l[MAX_N+1], small_r[MAX_N+1], last[MAX_N], join[MAX_N];
v<int> comp[MAX_N];
ll ans[MAX_N+1];
ll fac[MAX_N+1], inv[MAX_N+1];
void calc() {
fac[0] = 1;
for (int i=1;i<MAX_N+1;i++) {fac[i] = mul(fac[i-1],i);}
inv[MAX_N] = exp(fac[MAX_N],mod-2);
for (int i=MAX_N;i;i--) {inv[i-1] = mul(inv[i],i);}
}
ll C(int n, int k) {
return mul(fac[n],mul(inv[k],inv[n-k]));
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
calc();
int n, k; cin >> n >> k;
stack<int> s, s2;
for (int i=0;i<n;i++) {
cin >> a[i]; a[i]--;
}
for (int i=0;i<n;i++) {
while (s.size() && a[s.top()]>=a[i]) {s.pop();}
small_l[i] = (s.empty() ? -1 : s.top());
while (s2.size() && a[s2.top()]>=a[n-i-1]) {s2.pop();}
small_r[n-i-1] = (s2.empty() ? n : s2.top());
s.push(i); s2.push(n-i-1);
last[a[i]] = n;
ans[i] = 1;
join[i] = -1;
}
small_l[n] = n+1; small_r[n] = 0;
for (int i=n-1;i>=0;i--) {
if (i<small_l[last[a[i]]]) {
if (small_r[i]==small_l[last[a[i]]]) {join[last[a[i]]] = i;}
last[a[i]] = i;
}
comp[last[a[i]]].pb(i);
}
for (int i=0;i<n;i++) {
if (comp[i].empty()) {continue;}
reverse(all(comp[i]));
int m = comp[i].size();
if (join[i]!=-1) {
int tot = m+comp[join[i]].size();
mul2(ans[small_l[i]],C(tot,m));
mul2(ans[small_l[i]+1],exp(C(tot,m),mod-2));
}
for (int j=0;j<m;j++) {
mul2(ans[comp[i][j]],C(m-1,j));
mul2(ans[comp[i][j]+1],exp(C(m-1,j),mod-2));
if (j && comp[i][j]-1 != comp[i][j-1]) {
mul2(ans[comp[i][j-1]+1],C(m,j));
mul2(ans[comp[i][j]],exp(C(m,j),mod-2));
}
}
}
for (int i=0;i<n;i++) {
mul2(ans[i+1],ans[i]);
}
while (k--) {
int p; cin >> p;
cout << ans[p-1] << '\n';
}
/*
assume this light is left of range
then find nearest strictly small_ler value
you can turn of until there,
if an L and R range of same energy lightbulbs intersect, then any
point inside them can reach this state, and has two options from here
(biconditional)
Find all components of L R intersection, then calc for each point and subrange
in the component, with prefix multiplication/mod inverse division,
and then sweep through once to get answer
question: how to calc answer for a component
have L points on left, R points on right
it should be (L+R) choose L
this means that we only need nearest small_ler values on either R or L
*/
}
