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;
const int N = 500005;
int n, m, k, ne[N], f[N], g[N];
char s[N], t[N];
signed main() {
  scanf("%d%d%d", &n, &m, &k);
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  for (int i = 2, j = 0; i <= m; i++) {
    while (j && (j == k || t[j + 1] != t[i])) j = ne[j];
    if (t[j + 1] == t[i]) j++;
    ne[i] = j;
  }
  memset(f, 0x3f, sizeof f);
  for (int i = 1, j = 0; i <= n; i++) {
    while (j && (j == k || t[j + 1] != s[i])) j = ne[j];
    if (t[j + 1] == s[i]) j++;
    if (j == m) {
      puts("Yes");
      int l = max(1, i - k - k + 1);
      printf("%d %d\n", l, l + k);
      return 0;
    }
    if (i >= k) f[j] = min(f[j], i - k + 1);
  }
  for (int i = m; i; i--) f[ne[i]] = min(f[ne[i]], f[i]);
  reverse(s + 1, s + n + 1);
  reverse(t + 1, t + m + 1);
  for (int i = 2, j = 0; i <= m; i++) {
    while (j && (j == k || t[j + 1] != t[i])) j = ne[j];
    if (t[j + 1] == t[i]) j++;
    ne[i] = j;
  }
  for (int i = 1, j = 0; i <= n; i++) {
    while (j && (j == k || t[j + 1] != s[i])) j = ne[j];
    if (t[j + 1] == s[i]) j++;
    if (i >= k) g[j] = max(g[j], n - i + 1);
  }
  for (int i = m; i; i--) g[ne[i]] = max(g[ne[i]], g[i]);
  for (int i = 1; i <= m; i++)
    if (f[i] + k <= g[m - i]) {
      puts("Yes");
      printf("%d %d\n", f[i], g[m - i]);
      return 0;
    }
  puts("No");
}
