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[501000];
char t[501000];
int p[501000];
int n, m, k;
int a[501000];
int b[501000];
void calc(int i, int j, int *a) {
  while (a[j] == 0) {
    a[j] = i;
    j = p[j];
  }
}
void work(int *a) {
  int j = 0;
  for (int i = 2; i <= m; i++) {
    while (j && t[j + 1] != t[i]) j = p[j];
    if (t[j + 1] == t[i]) j++;
    p[i] = j;
  }
  j = 0;
  for (int i = 1; i <= n; i++) {
    while (j && t[j + 1] != s[i]) j = p[j];
    if (t[j + 1] == s[i]) j++;
    if (i >= k) calc(i, j, a);
    if (j == m && i < k) a[j] = k;
    if (j == m) j = p[j];
  }
}
int main() {
  cin >> n >> m >> k;
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  work(a);
  for (int i = 1; i <= n / 2; i++) swap(s[i], s[n - i + 1]);
  for (int i = 1; i <= m / 2; i++) swap(t[i], t[m - i + 1]);
  work(b);
  for (int i = 0; i <= min(k, m); i++) {
    if (a[i] && b[m - i] && m - i <= k)
      if (a[i] - k + 1 + k <= n - b[m - i] + 1) {
        cout << "Yes" << endl;
        cout << a[i] - k + 1 << " " << n - b[m - i] + 1 << endl;
        return 0;
      }
  }
  cout << "No" << endl;
}
