Jenya has recently acquired quite a useful tool — k-scissors for cutting strings. They are generally used for cutting out two non-intersecting substrings of length k from an arbitrary string s (its length should be at least 2·k in order to perform this operation) and concatenating them afterwards (preserving the initial order). For example, with the help of 2-scissors you can cut ab and de out of abcde and concatenate them into abde, but not ab and bc since they're intersecting.

It's a nice idea to test this tool before using it in practice. After looking through the papers, Jenya came up with two strings s and t. His question is whether it is possible to apply his scissors to string s such that the resulting concatenation contains t as a substring?

Input

The first line contains three integers n, m, k (2 ≤ m ≤ 2·k ≤ n ≤ 5·105) — length of s, length of t and the aforementioned scissors' parameter correspondingly.

The next two lines feature s and t consisting of lowercase latin letters.

Output

If there is no answer, print «No». 

Otherwise print «Yes» and two integers L and R denoting the indexes where cutted substrings start (1-indexed). If there are several possible answers, output any.

Examples

Input

7 4 3
baabaab
aaaa


Output

Yes
1 5


Input

6 3 2
cbcbcb
bcc


Output

Yes
2 5


Input

7 5 3
aabbaaa
aaaaa


Output

No

Note

In the first sample case you can cut out two substrings starting at 1 and 5. The resulting string baaaab contains aaaa as a substring.

In the second sample case the resulting string is bccb.
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXLEN = 500000;
int slen, tlen, sz;
char s[MAXLEN + 1];
char t[MAXLEN + 1];
int pref[MAXLEN + 1], fst;
int suff[MAXLEN + 1], lst;
int nxt[MAXLEN + 1];
void calc(int *res, int &fst) {
  nxt[0] = nxt[1] = 0;
  for (int i = (2); i <= (tlen); ++i) {
    int cur = nxt[i - 1];
    while (cur > 0 && t[cur] != t[i - 1]) cur = nxt[cur];
    if (t[cur] == t[i - 1]) ++cur;
    nxt[i] = cur;
  }
  int at = 0;
  for (int i = (0); i <= (tlen); ++i) res[i] = -1;
  fst = -1;
  if (at >= sz) res[at] = 0;
  for (int i = (0); i < (slen); ++i) {
    while (at > 0 && t[at] != s[i]) at = nxt[at];
    if (t[at] == s[i]) ++at;
    if (i + 1 >= sz && res[at] == -1) res[at] = i + 1;
    if (at == tlen && fst == -1) fst = i + 1;
    if (at == tlen) at = nxt[at];
  }
  for (int i = tlen; i > 0; --i)
    if (res[i] != -1 && (res[nxt[i]] == -1 || res[i] < res[nxt[i]]))
      res[nxt[i]] = res[i];
}
void run() {
  scanf("%d%d%d", &slen, &tlen, &sz);
  scanf("%s", s);
  scanf("%s", t);
  calc(pref, fst);
  reverse(s, s + slen);
  reverse(t, t + tlen);
  calc(suff, lst);
  reverse(s, s + slen);
  reverse(t, t + tlen);
  reverse(suff, suff + tlen + 1);
  for (int i = (0); i <= (tlen); ++i)
    if (suff[i] != -1) suff[i] = slen - suff[i];
  if (lst != -1) lst = slen - lst;
  if (fst != -1 && fst < sz || lst != -1 && lst > slen - sz) {
    printf("Yes\n%d %d\n", 1, slen - sz + 1);
    return;
  }
  for (int i = (0); i <= (tlen); ++i)
    if (pref[i] != -1 && suff[i] != -1 && pref[i] <= suff[i] && i <= sz &&
        tlen - i <= sz) {
      printf("Yes\n%d %d\n", pref[i] - sz + 1, suff[i] + 1);
      return;
    }
  printf("No\n");
}
int main() {
  run();
  return 0;
}
