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 inf = 0x3fffffff;
const int maxn = int(5e5) + 100;
int lenst, lenstr, cut;
char st[maxn], str[maxn];
int le[maxn], ri[maxn], fail[maxn];
int minle[maxn], maxri[maxn];
void read() {
  scanf("%d%d%d%*c", &lenst, &lenstr, &cut);
  fgets(st + 1, lenst + 2, stdin);
  fgets(str + 1, lenstr + 2, stdin);
}
void KMP(int *match, int *limmatch) {
  for (int i = 2, j = 0; i <= lenstr; ++i) {
    while (j != 0 && str[j + 1] != str[i]) j = fail[j];
    if (str[j + 1] == str[i]) ++j;
    fail[i] = j;
  }
  for (int i = 1; i <= lenstr; ++i) match[i] = limmatch[i] = inf;
  for (int i = 1, j = 0; i <= lenst; ++i) {
    while (j != 0 && str[j + 1] != st[i]) j = fail[j];
    if (str[j + 1] == st[i]) ++j;
    if (i >= cut) match[j] = min(match[j], i);
    limmatch[j] = min(limmatch[j], i);
    if (j == lenstr) j = fail[j];
  }
  for (int i = lenstr; i; --i) match[fail[i]] = min(match[fail[i]], match[i]);
}
void solve() {
  KMP(le, minle);
  for (int i = 1; i <= lenstr / 2; ++i) swap(str[i], str[lenstr - i + 1]);
  for (int i = 1; i <= lenst / 2; ++i) swap(st[i], st[lenst - i + 1]);
  KMP(ri, maxri);
  for (int i = 1; i <= lenstr; ++i) {
    ri[i] = lenst - ri[i] + 1;
    maxri[i] = lenst - maxri[i] + 1;
  }
  for (int i = 1; i <= lenstr / 2; ++i) {
    swap(ri[i], ri[lenstr - i + 1]);
    swap(maxri[i], maxri[lenstr - i + 1]);
  }
  if (cut >= lenstr && minle[lenstr] < inf) {
    int L = max(1, minle[lenstr] - cut + 1);
    if (L + cut - 1 + cut <= lenst) {
      puts("Yes");
      printf("%d %d\n", L, L + cut);
      return;
    }
    int R = min(lenst, maxri[1] + cut - 1);
    if (R - cut + 1 - cut > 0) {
      puts("Yes");
      printf("%d %d\n", R - cut - cut + 1, R - cut + 1);
      return;
    }
  }
  le[0] = cut;
  ri[lenstr + 1] = lenst - cut + 1;
  for (int i = 0; i <= lenstr; ++i)
    if (le[i] < ri[i + 1] && i <= cut && lenstr - i <= cut) {
      puts("Yes");
      printf("%d %d\n", le[i] - cut + 1, ri[i + 1]);
      return;
    }
  puts("No");
}
int main() {
  read();
  solve();
  return 0;
}
