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>
#pragma GCC optimize("O3")
using namespace std;
const int mod = 1000000007;
const int N = 500005;
void Getnext(string &s, int *fail) {
  int id = -1;
  fail[0] = -1;
  for (int i = 1; i < s.size(); ++i) {
    while (~id && s[id + 1] != s[i]) id = fail[id];
    if (s[id + 1] == s[i]) ++id;
    fail[i] = id;
  }
}
int A[N], B[N];
int kmp1[N], kmp2[N];
int lf[N], ri[N];
int main() {
  int n, m, k;
  string s, t1, t2;
  cin >> n >> m >> k;
  cin >> s >> t1;
  t2 = t1;
  reverse(t2.begin(), t2.end());
  s.push_back('$');
  t1.push_back('$');
  t2.push_back('$');
  Getnext(t1, A);
  Getnext(t2, B);
  memset(lf, -1, sizeof(lf));
  memset(ri, -1, sizeof(ri));
  lf[0] = k - 1;
  ri[0] = n - k;
  for (int i = 0, j = -1; i < n; i++) {
    while (j != -1 && s[i] != t1[j + 1]) j = A[j];
    if (s[i] == t1[j + 1]) j++;
    if (j == m - 1 && i < k - 1) {
      puts("Yes");
      printf("%d %d\n", 1, 1 + k);
      return 0;
    }
    if (i < k - 1) continue;
    int pos = j;
    while (pos != -1) {
      if (lf[pos + 1] != -1) break;
      lf[pos + 1] = i;
      pos = A[pos];
    }
    if (j == m - 1 && i - k + 1 >= k) {
      puts("Yes");
      printf("%d %d\n", i - k + 1 + 1 - k, i - k + 1 + 1);
      return 0;
    }
  }
  for (int i = n - 1, j = -1; i >= 0; i--) {
    while (j != -1 && s[i] != t2[j + 1]) j = B[j];
    if (s[i] == t2[j + 1]) j++;
    if (j == m - 1 && i > n - k) {
      puts("Yes");
      printf("%d %d\n", n - 2 * k + 1, n - k + 1);
      return 0;
    }
    if (i > n - k) continue;
    int pos = j;
    while (pos != -1) {
      if (ri[pos + 1] != -1) break;
      ri[pos + 1] = i;
      pos = B[pos];
    }
    if (j == m - 1 && n - i - k >= k) {
      puts("Yes");
      printf("%d %d\n", i + 1, i + k + 1);
      return 0;
    }
  }
  for (int i = 0; i <= m; i++) {
    if (lf[i] == -1 || ri[m - i] == -1) continue;
    if (i > k || m - i > k) continue;
    if (lf[i] < ri[m - i]) {
      puts("Yes");
      printf("%d %d\n", lf[i] - k + 1 + 1, ri[m - i] + 1);
      return 0;
    }
  }
  puts("No");
}
