#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end()
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;
template<template<typename> class Container, typename T>
ostream& operator<<(ostream& os, Container<T> o) {
os << "{";
int g = size(o);
for (auto i : o) os << i << ((--g) == 0 ? "" : ", ");
os << "}";
return os;
}
void _print() {
cerr << "\n";
}
template<typename T, typename ...V>
void _print(T t, V... v) {
cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...);
}
#ifdef LOCAL
#define dbg(x...) cerr << #x << " = "; _print(x);
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif
/*
try sus kmp?
precompute the indices i and j for each k in the pattern
to quickly check is some building can be glued to the end of the current match
a match is valid if and only if this is satisfied for all buildings
compute a fallback array, such that if a mismatch happens on index i you can delete
the first x amount to make the prefix match again
compute an auxilliary array dp, where dp[i] = how far away the next mismatch is
if you were to start matching at index i
if you know that from index i, the first mismatched character is j, and that
if you were to match with yourself from index 1 the next mismatched character is at 1 + k,
then
case 1: at index i + 1, you know for certain that index i + 1 + k is a mismatch, and i + 1 + k < j
this means that you know the next mismatched position
you can continue doing this until case 2
case 2: at index i + 1, you know for certain that index i + 1 + k is a mismatch, and i + 1 + k >= j
who knows, it might match now!
start checking again from index i + 1
*/
int m, n;
vt<int> pat;
vt<pi> comp;
vt<int> h;
// check match at index j of the pattern starting at index i in str
bool match(int i, int j, vt<int>& str) {
// cerr << "matching " << j << " starting from " << i << endl;
pi prv = comp[j - i];
bool good = true;
if (prv.f != -1) good &= str[prv.f + i] < str[j];
if (prv.s != -1) good &= str[j] < str[prv.s + i];
return good;
}
vt<int> dp;
void compute_dp() {
dp.resize(m, inf);
dp[0] = 0;
int j = 1; // the first mismatched character starting from index i
FOR (i, 1, m) {
while (j < m && match(i, j, pat)) j++;
cerr << "matched from " << i << " up to " << j << endl;
dp[i] = j - i;
cerr << "dp " << i << " = " << j - i << endl;
int k = 1;
cerr << i + k << " " << (i + k < m ? i + k + dp[k] : -1) << endl;
while (i + k < m && i + k + dp[k] < j) { // case 1: we know the mismatch happens in our good prefix
cerr << "we know the next mismatch\n";
dp[i + k] = dp[k];
cerr << "dp " << i + k << " = " << dp[k] << endl;
k++;
}
i += k - 1;
// case 2: idk whats happenign anymore
}
dbg(dp);
}
vt<int> fb;
void compute_fallback() {
fb.resize(m + 1);
fb[0] = -1;
vt<int> mod_dp = dp; // convert from next mismatch to last matching index relative to start
mod_dp[0] = 1;
dbg(mod_dp);
vt<int> last(m + 1, m);
ROF (i, 0, m) {
fb[i + 1] = last[mod_dp[i]] - i;
last[mod_dp[i]] = i;
dbg(last);
}
dbg(fb);
}
main() {
cin.tie(0)->sync_with_stdio(0);
cin >> m >> n;
pat.resize(m);
F0R (i, m) {
int v; cin >> v;
pat[v - 1] = i;
}
comp.resize(m, {-1, -1});
map<int, int> prv;
F0R (i, m) {
auto it = prv.lower_bound(pat[i]);
if (it != prv.end()) comp[i].s = it->s;
if (it != prv.begin()) comp[i].f = prev(it)->s;
prv[pat[i]] = i;
}
h.resize(n);
F0R (i, n) cin >> h[i];
dbg(pat);
F0R (i, m) {
cerr << comp[i].f << " " << comp[i].s << endl;
}
compute_dp();
compute_fallback();
vt<int> out;
int i = 0, j = 0;
while (i < n) {
while (j < n && j - i < m && match(i, j, h)) j++;
cerr << "starting from " << i << ", mismatched at " << j << endl;
// j is now at first mismatch
if (j - i == m) out.pb(i);
i += fb[j - i];
cerr << "i = " << i << endl;
}
cout << size(out) << endl;
for (int e : out) cout << e + 1 << " ";
cout << endl;
}
