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;
long long MOD = 998244353;
long long base = 131;
int n, m, k;
char s[500010], t[500010];
struct hash {
  long long hash[500010], pw[500010];
  void hs(char *s) {
    int len = strlen(s + 1);
    pw[0] = 1;
    for (int i = 1; i <= len; ++i) {
      hash[i] = (hash[i - 1] * base + s[i] - 'a') % MOD;
      pw[i] = pw[i - 1] * base % MOD;
    }
  }
  long long get(int l, int r) {
    return ((hash[r] - hash[l - 1] * pw[r - l + 1]) % MOD + MOD) % MOD;
  }
} S, T;
int maxl[500010], maxr[500010];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  scanf("%s%s", s + 1, t + 1);
  S.hs(s), T.hs(t);
  int pos = k;
  for (int i = 1; i <= m; ++i) maxl[i] = n + 1;
  for (int i = 1; i <= min(m, k); ++i) {
    while (pos <= n && S.get(pos - i + 1, pos) != T.get(1, i)) ++pos;
    if (S.get(k - i + 1, k) == T.get(1, i)) pos = k;
    maxl[i] = pos;
  }
  pos = n - k + 1;
  for (int i = 1; i <= min(m, k); ++i) {
    while (pos && S.get(pos, pos + i - 1) != T.get(m - i + 1, m)) --pos;
    if (S.get(n - k + 1, n - k + i) == T.get(m - i + 1, m)) pos = n - k + 1;
    maxr[m - i + 1] = pos;
  }
  for (int i = 1; i <= n - m + 1; ++i) {
    if (S.get(i, i + m - 1) == T.get(1, m)) {
      if (k >= i && n - k + 1 <= i + m - 1) continue;
      printf("Yes\n");
      return printf("%d %d\n", min(max(1, i - k + 1), n - k + 1 - k),
                    max(k + 1, min(n - k + 1, i))),
             0;
    }
  }
  for (int i = 1; i < m; ++i) {
    if (maxl[i] < maxr[i + 1] && maxl[i] <= n && maxr[i + 1]) {
      printf("Yes\n");
      return printf("%d %d\n", maxl[i] - k + 1, maxr[i + 1]), 0;
    }
  }
  printf("No\n");
}
