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[500001], t[500001];
int failure[500001], prefixFailure[500001], suffixFailure[500001],
    leftEnd[500001], maxPrefixFailure[500001];
void ReverseString(char s[], int n = -1) {
  n = (n >= 0) ? n : strlen(s);
  for (int i = 0; 2 * i < n; ++i) swap(s[i], s[n - 1 - i]);
}
int ComputeFailureValue(int j, int k, char c) {
  while ((j >= k) || (j != 0 && t[j] != c)) {
    j = failure[j];
  }
  j += (t[j] == c);
  return j;
}
void ComputeFailureFunction(char s[], int failureS[], int n, int k = 500001,
                            bool isPattern = 1) {
  for (int i = 1 + isPattern, j = 0; i <= n; ++i) {
    char c = s[i - 1];
    j = ComputeFailureValue(j, k, c);
    failureS[i] = j;
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  scanf("%s%s", s, t);
  ComputeFailureFunction(t, failure, m);
  ComputeFailureFunction(s, prefixFailure, n, n, 0);
  for (int i = 1; i <= n; ++i) {
    if (prefixFailure[i] == m) {
      int j = max(1, i - 2 * k + 1);
      printf("Yes\n%d %d\n", j, j + k);
      return 0;
    }
  }
  ComputeFailureFunction(s, prefixFailure, n, k, 0);
  ReverseString(t, m);
  ComputeFailureFunction(t, failure, m);
  for (int i = 2; i <= n; ++i)
    maxPrefixFailure[i] = max(prefixFailure[i], maxPrefixFailure[i - 1]);
  for (int i = n, j = 0; i > 0; --i) {
    j = ComputeFailureValue(j, m - maxPrefixFailure[i - 1], s[i - 1]);
    suffixFailure[i] = j;
  }
  for (int i = 1, j = 0; i <= n; ++i) {
    if (j + suffixFailure[i] >= m && leftEnd[m - suffixFailure[i]] >= k &&
        i + k - 1 <= n) {
      printf("Yes\n%d %d\n", leftEnd[m - suffixFailure[i]] - k + 1, i);
      return 0;
    }
    leftEnd[prefixFailure[i]] = i;
    j = max(j, prefixFailure[i]);
  }
  printf("No\n");
  return 0;
}
