对每个字符，做一遍匹配的NTT, 即设每个位置能匹配的值为f_i, 那么f_i = s_i + j == t_j(0 < j < t.size() )，然后每个字符都匹配一下，加到一起，暴力枚举对于位置i, 是否存在f_i + k >= t.size()就行

我的ntt板子常数大，换个常数小或者用fft应该就能过//我直接玄学剪枝跑出来的

#include <bits/stdc++.h>
#define endl '\n'
#define pb push_back

using namespace std;
using ll = long long;
using db = double;
using pii = pair<int , int>;

const int maxn  = 4e6 + 10;

int zm[220][2];
const int mod = 998244353;
const int G = 3;
int n,m,L,R[maxn];
int A[maxn],B[maxn];
ll qpow(ll a,ll b)
{
    ll ans = 1;
    while(b>0)
    {
    	if(b&1) ans = ans*a%mod;
    	b>>=1;
    	a = a*a%mod;
	}
	return ans%mod;
}
void NTT(int *a,int f)
{
    for(int i = 0; i < n; i++) 
	{
		if(i < R[i]) swap(a[i],a[R[i]]);
	}
    for(int i = 1; i < n; i <<= 1)
	{
        ll gn = qpow(G,(mod - 1) / (i << 1));
        for(int j = 0; j < n; j += (i << 1))
		{
            ll g = 1;
            for(int k = 0; k < i; k++, g = g * gn % mod)
			{
                int x = a[j + k], y = g * a[j + k + i] % mod;
                a[j + k] = (x + y) % mod; 
				a[j + k + i] = (x - y + mod) % mod;
            }
        }
    }
    if(f==1) return;
    int inv = qpow(n,mod - 2); 
	reverse(a + 1,a + n);
    for(int i = 0; i < n; i++) a[i] = 1ll * a[i] * inv % mod;
}
void solve(int *A,int *B)
{
	m = n + m;
	L = 0;
	for(n = 1; n <= m; n <<= 1) L++;
    for(int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	NTT(A,1); 
	NTT(B,1);
    for(int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % mod;
    NTT(A,-1);
}
int ans[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int i, j, k;
	cin >> k;
	string s, t;
	cin >> s >> t;
	for (auto c : s)
		zm[c][0]++;
	for (auto c : t)
		zm[c][1]++;
	int ty = 0;
	for (i = 0; i < 220; i++)
	{
		if (zm[i][0] && zm[i][1])
		{
			for (j = 0; j < n; j++) A[j] = B[j] = 0;
			n = s.size(), m = t.size();
			for (j = 0; j < n; j++)
				A[j] = s[j] == i;
			for (j = 0; j < m; j++)
				B[m - 1 - j] = t[j] == i;
			solve(A, B);
			int nn = s.size(), mm = t.size();
			for (int t = mm - 1; t < nn; t++)
				ans[t - mm + 1] += A[t];
			ty += 20 * nn;
		}
		if (ty > 80000000)
		 break;
	}
	bool flag = 0;
    for (i = 0; i < s.size(); i++)
    	if (ans[i] + k >= t.size())
    		flag = 1, cout << i + 1 << ' ';
    if (!flag)
    	cout << -1;
    return 0;
}