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 = 5e5 + 5;
char a[N], b[N];
int n, m, K, fail[N], pre[N], aft[N];
void cul(char *A, char *B, int *res) {
  int i, j;
  for (i = 2, j = 0; i <= m; i++) {
    while (j && B[j + 1] != B[i]) j = fail[j];
    if (B[j + 1] == B[i]) j++;
    fail[i] = j;
  }
  res[0] = K;
  for (i = 1; i <= m; i++) res[i] = n + 1;
  for (i = 1, j = 0; i <= n; i++) {
    while (j && B[j + 1] != A[i]) j = fail[j];
    if (B[j + 1] == A[i]) j++;
    if (i >= K) res[j] = min(res[j], i);
    if (j == m) {
      puts("Yes");
      int r = i, l = i - m + 1;
      while (r - l + 1 < 2 * K)
        if (l > 1)
          l--;
        else if (r < n)
          r++;
      cout << l << ' ' << r - K + 1, exit(0);
    }
  }
  for (i = 1; i <= m; i++) res[fail[i]] = min(res[fail[i]], res[i]);
}
int main() {
  cin >> n >> m >> K >> a + 1 >> b + 1;
  cul(a, b, pre);
  reverse(a + 1, a + n + 1), reverse(b + 1, b + m + 1);
  cul(a, b, aft);
  for (int i = 0; i <= m; i++) aft[i] = n - aft[i] + 1;
  for (int i = max(0, m - K); i <= min(K, m); i++)
    if (pre[i] < aft[m - i])
      return cout << "Yes" << endl << pre[i] - K + 1 << ' ' << aft[m - i], 0;
  puts("No");
}
