// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define sz(x) int(x.size())
#define pb push_back
using ll = long long;
template<class T> using V = vector<T>;
const int MOD1 = ll(1e9)+7;
// const int MOD2 = 998244353;
const int base = int(1e6) + 1007;
const int nax = (1 << 20);
int POW1[nax];
// int POW2[nax];
struct T {
int H1,/* H2,*/ len;
bool operator==(T a) {
return a.H1 == H1/* && a.H2 == H2*/;
};
};
T c;
T comb(T a, T b) {
c = a; c.len += b.len;
c.H1 = ((a.H1 * 1LL * POW1[b.len]) + b.H1) % MOD1;
// c.H2 = ((a.H2 * 1LL * POW2[b.len]) + b.H2) % MOD2;
return c;
}
// SEGTREE
T seg[2*nax];
void pull(int x) { seg[x] = comb(seg[2*x], seg[2*x+1]); }
void upd(int p, T v) { seg[p += nax] = v; for(p /= 2; p; p /= 2) pull(p); }
T get() { return seg[1]; }
int main() {
cin.tie(0)->sync_with_stdio(0);
POW1[0] = 1;
// POW2[0] = 1;
for(int t = 1; t < nax; t++) {
POW1[t] = (POW1[t - 1] * 1LL * base) % MOD1;
// POW2[t] = (POW2[t - 1] * 1LL * base) % MOD2;
}
for(int t = 0; t < 2*nax; t++) seg[t] = T{0, 0};
int N, M; cin >> N >> M;
V<int> A(N); for(auto& x : A) { cin >> x; --x; }
V<int> B(M); for(auto& x : B) cin >> x;
{
map<int, int> mp; int cur = 0;
for(auto& x : B) mp[x]++;
for(auto& p : mp) p.second = cur++;
for(auto& x : B) x = mp[x];
}
T H = {0, 0}, R = {0, 0}, REM = {0, 0};
for(int i = 0; i < N; i++) {
H = comb(H, {A[i] + 1, 1});
R = comb(R, {1, 1});
}
V<int> ans;
for(int i = 0; i < N; i++) upd(B[i], {i + 1, 1});
auto check = [&](int i) {
T res = get();
if ((res.H1 -= REM.H1) < 0) res.H1 += MOD1;
// if ((res.H2 -= REM.H2) < 0) res.H2 += MOD2;
if (res == H) ans.push_back(i);
};
check(0);
for(int i = N; i < M; i++) {
if ((REM.H1 += R.H1) >= MOD1) REM.H1 -= MOD1;
// if ((REM.H2 += R.H2) >= MOD2) REM.H2 -= MOD2;
upd(B[i], {i + 1, 1});
upd(B[i - N], {0, 0});
check(i - N + 1);
}
cout << sz(ans) << nl;
for(auto& x : ans) cout << x+1 << " ";
cout << nl;
return 0;
}
