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;
int nxt[500100], id[500100];
char s[500100], t[500100];
int L[500100], R[500100];
void init(char *s) {
  nxt[0] = -1;
  int len = strlen(s);
  int i = 0, j = -1;
  while (i < len) {
    if (j == -1 || s[i] == s[j]) {
      i++;
      j++;
      nxt[i] = j;
    } else {
      j = nxt[j];
    }
  }
}
int kmp(char *s, char *t, int l) {
  int lens = strlen(s), lent = strlen(t);
  int r = lent - 1;
  int j = 0, i = 0;
  while (i < lens) {
    if (j == -1 || s[i] == t[j]) {
      if (j != -1 && i >= l && id[j] == -1) r = min(r, j - 1), id[j] = i;
      i++;
      j++;
    } else {
      if (i - 1 >= l) {
        for (int o = max(0, j - (i - l)); o <= min(j - 1, r); o++)
          if (id[o] == -1) id[o] = i - (j - o);
        r = min(r, max(0, j - (i - l)) - 1);
      }
      j = nxt[j];
    }
    if (j == lent) return i;
  }
  return -1;
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  scanf("%s", s);
  scanf("%s", t);
  init(t);
  memset(id, -1, sizeof(id));
  int ok = kmp(s, t, k - 1);
  if (ok != -1) {
    for (int i = 0; i < n; i++)
      if (i + k + k >= ok) return 0 * printf("Yes\n%d %d", i + 1, i + k + 1);
  }
  for (int i = 0; i < m; i++) L[i] = id[i];
  for (int i = 0; i < (n >> 1); i++) swap(s[i], s[n - i - 1]);
  for (int i = 0; i < (m >> 1); i++) swap(t[i], t[m - i - 1]);
  memset(nxt, -1, sizeof(nxt));
  memset(id, -1, sizeof(id));
  init(t);
  kmp(s, t, k - 1);
  for (int i = 0; i < m; i++) R[m - i - 1] = (id[i] == -1 ? -1 : n - id[i] - 1);
  for (int i = max(0, m - k - 1); i < min(k, m - 1); i++) {
    if (L[i] != -1 && R[i + 1] != -1 && R[i + 1] > L[i])
      return 0 * printf("Yes\n%d %d", L[i] - k + 2, R[i + 1] + 1);
  }
  return 0 * puts("No");
}
