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;
char s[500000 + 5], t[500000 + 5];
int n, m, k, nx[500000 + 5], a[500000 + 5], b[500000 + 5];
void cal(int* a) {
  int i, j;
  for (i = 2, j = 0; i <= m && i <= k; nx[i++] = j) {
    while (j && t[j + 1] != t[i]) j = nx[j];
    if (t[j + 1] == t[i]) ++j;
  }
  for (i = 1, j = 0; i <= n; ++i) {
    while (j && (j == k || t[j + 1] != s[i])) j = nx[j];
    if (t[j + 1] == s[i]) ++j;
    if (j == m) {
      if (i < 2 * k)
        printf("Yes\n1 %d", k + 1);
      else
        printf("Yes\n%d %d", i - k - k + 1, i - k + 1);
      exit(0);
    }
    if (i - k + 1 > 0) a[j] = min(a[j], i);
  }
  for (i = m; i; --i) a[nx[i]] = min(a[nx[i]], a[i]);
}
int main() {
  scanf("%d%d%d%s%s", &n, &m, &k, s + 1, t + 1);
  memset(a, 40, sizeof(a));
  cal(a);
  for (int i = 1; i < n + 1 - i; ++i) swap(s[i], s[n + 1 - i]);
  for (int i = 1; i < m + 1 - i; ++i) swap(t[i], t[m + 1 - i]);
  memset(b, 40, sizeof(b));
  cal(b);
  for (int i = 0; i <= m; ++i)
    if (a[i] <= n && b[m - i] <= n && a[i] < n - b[m - i] + 1)
      return 0 * printf("Yes\n%d %d", a[i] - k + 1, n - b[m - i] + 1);
  puts("No");
}
