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 maxn = 500010;
int n, m, k;
int kmp1[maxn], kmp2[maxn];
int len1[maxn], len2[maxn];
int hah[maxn];
char s[maxn], t[maxn];
template <typename Tp>
void read(Tp &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
int main() {
  read(n), read(m), read(k);
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  int j = 0;
  for (int i = 2; i <= m; i++) {
    while (j && t[j + 1] != t[i]) j = kmp1[j];
    if (t[j + 1] == t[i]) j++;
    kmp1[i] = j;
  }
  j = 0;
  for (int i = 1; i <= n; i++) {
    while (j && t[j + 1] != s[i]) j = kmp1[j];
    if (t[j + 1] == s[i]) j++;
    len1[i] = j;
    if (j == m) {
      int ps = j - m + 1;
      if (ps + 2 * k - 1 <= n) printf("Yes\n%d %d\n", ps, ps + k);
      return 0;
    }
    if (len1[i] > k || i - k < 0) len1[i] = -1;
  }
  j = 0;
  reverse(s + 1, s + n + 1);
  reverse(t + 1, t + m + 1);
  for (int i = 2; i <= m; i++) {
    while (j && t[j + 1] != t[i]) j = kmp2[j];
    if (t[j + 1] == t[i]) j++;
    kmp2[i] = j;
  }
  j = 0;
  for (int i = 1; i <= n; i++) {
    while (j && t[j + 1] != s[i]) j = kmp2[j];
    if (t[j + 1] == s[i]) j++;
    len2[i] = j;
    if (len2[i] > k || n - i + 1 + k > n + 1) len2[i] = -1;
  }
  reverse(len2 + 1, len2 + n + 1);
  int ps = 0;
  for (int i = 1; i <= n; i++) {
    if (len2[i] == -1) continue;
    int x = len2[i];
    if (n < 499999) {
      while (x) {
        if (hah[m - x]) {
          ps = i;
          break;
        }
        x = kmp2[x];
      }
    } else if (hah[m - x] || hah[m - kmp2[x]] || hah[m - kmp2[kmp2[x]]]) {
      ps = i;
      break;
    }
    if (len1[i] != -1) hah[len1[i]]++;
  }
  if (!ps)
    printf("No\n");
  else {
    for (int i = 1; i < ps; i++) {
      int x = len2[ps];
      if (n < 499999) {
        while (x) {
          if (len1[i] == m - x) {
            printf("Yes\n%d %d\n", i - k + 1, ps);
            return 0;
          }
          x = kmp2[x];
        }
      } else if (len1[i] == m - x || len1[i] == m - kmp2[x] ||
                 len1[i] == m - kmp2[kmp2[x]]) {
        printf("Yes\n%d %d\n", i - k + 1, ps);
        return 0;
      }
    }
  }
  return 0;
}
