Encoding

Polycarp invented a new way to encode strings. Let’s assume that we have string T, consisting of lowercase English letters. Let’s choose several pairs of letters of the English alphabet in such a way that each letter occurs in at most one pair. Then let’s replace each letter in T with its pair letter if there is a pair letter for it. For example, if you chose pairs (l, r), (p, q) and (a, o), then word “parallelogram” according to the given encoding principle transforms to word “qolorreraglom”.

Polycarpus already has two strings, S and T. He suspects that string T was obtained after applying the given encoding method from some substring of string S. Find all positions m _i in S (1 ≤ m _i ≤ |S| - |T| + 1), such that T can be obtained fro substring S _m i S _m i + 1… S _{m i + |T| - 1} by applying the described encoding operation by using some set of pairs of English alphabet letters

Input

The first line of the input contains two integers, |S| and |T| (1 ≤ |T| ≤ |S| ≤ 2·10⁵) — the lengths of string S and string T, respectively.

The second and third line of the input contain strings S and T, respectively. Both strings consist only of lowercase English letters.

Output

Print number k — the number of suitable positions in string S.

In the next line print k integers m ₁, m ₂, …, m _k — the numbers of the suitable positions in the increasing order.

Examples

input

11 5
abacabadaba
acaba

output

3
1 3 7

input

21 13
paraparallelogramgram
qolorreraglom

output

1
5

Solution:

#include <bits/stdc++.h>

int const N = 234567;
int const X = 33533;
long long M1;
long long M2;  // TODO TODO CHANGE
long long const MASK = (1LL << 32) - 1;

long long POW1[N], POW2[N];

long long add1(long long h1, int c) {
  return (h1 * X + c) % M1;
}

long long add2(long long h1, int c) {
  return (h1 * X + c) % M2;
}

long long add(long long const & a, int c) {
  return (add1(a >> 32, c) << 32) | add2(a & MASK, c);
}

long long rem1(long long h1, int len) {
    h1 -= POW1[len];
    if (h1 < 0) h1 += M1;
    return h1;
}

long long rem2(long long h2, int len) {
    h2 -= POW2[len];
    if (h2 < 0) h2 += M2;
    return h2;
}

long long rem(long long const & a, int len) {
  return (rem1(a >> 32, len) << 32) | rem2(a & MASK, len);
}

int n;

std::mt19937 rnd(std::chrono::system_clock::now().time_since_epoch().count());

int const HALFBILLION = 500000000;

bool prime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}

void gen(long long & M) {
  M = rnd() % HALFBILLION + HALFBILLION;
  while (!prime(M)) {
    ++M;
  }
}

int s[N], t[N], ans[N];

long long tmp[42], h[42], sh[42];
int id1[42], id2[42], mp[42];

void read(int & n, int * s) {
    int c = getchar();
    while (c <= 32) c = getchar();
    n = 0;
    while (c > 32) {
s[n++] = c - 'a';
c = getchar();
}
}

bool check() {
std::sort(id2, id2 + 26, [](int i, int j) { return sh[i] < sh[j]; });
    for (int i = 0; i < 26; i++) mp[i] = -1;
    for (int it = 0; it < 26; it++) {
        int i = id1[it];
        int j = id2[it];
        if (h[i] != sh[j]) return false;
        if (h[i] == 0 && sh[j] == 0) continue;
        if (mp[i] >= 0 && mp[i] != j) return false;
if (mp[j] >= 0 && mp[j] != i) return false;
mp[i] = j;
mp[j] = i;
}
return true;
}

int main() {
gen(M1);
gen(M2);
POW1[0] = POW2[0] = 1;
for (int i = 1; i < N; i++) {
      POW1[i] = POW1[i - 1] * X % M1;
      POW2[i] = POW2[i - 1] * X % M2;
    }
    int n, m;
    scanf("%d%d", &n, &m);
    read(n, s);
    read(m, t);
    for (int i = 0; i < 26; i++) {
        h[i] = 0;
        for (int j = 0; j < m; j++) {
            h[i] = add(h[i], t[j] == i);
        }
    }
    for (int i = 0; i < 26; i++) id1[i] = id2[i] = i;
    std::sort(id1, id1 + 26, [](int i, int j) { return h[i] < h[j]; });
    for (int i = 0; i < 26; i++) sh[i] = 0;
    for (int i = 0; i < n && i < m; i++) {
        for (int j = 0; j < 26; j++) sh[j] = add(sh[j], s[i] == j);
    }
    int ac = 0;
    if (check()) {
        ans[ac++] = 0;
    }
    long long xor1 = 0;
    for (int i = 0; i < 26; i++) xor1 ^= h[i];
    for (int i = m; i < n; i++) {
        long long xor2 = 0;
        int torem = s[i - m];
        for (int j = 0; j < 26; j++) {
            sh[j] = add(sh[j], s[i] == j);
            if (j != torem) xor2 ^= sh[j];
        }
        sh[torem] = rem(sh[torem], m);
        xor2 ^= sh[torem];
        if (xor1 == xor2 && check()) {
            ans[ac++] = i - m + 1;
        }
    }
    printf("%d\n", ac);
    for (int i = 0; i < ac; i++) {
        if (i > 0) putchar(' ');
printf("%d", ans[i] + 1);
}
puts("");
}