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 N = 1e6 + 8;
int n, m, k, len, f[N], g[N], w1[N], w2[N];
char s[N], t[N], str[N];
void kmp(int* nxt) {
  nxt[1] = 0;
  for (int i = 2, j = 0; i <= len; i++) {
    while (j && str[i] != str[j + 1]) j = nxt[j];
    nxt[i] = str[i] == str[j + 1] ? (++j) : j;
  }
}
int main() {
  scanf("%d%d%d%s%s", &n, &m, &k, s + 1, t + 1);
  if (k * 2 > n || k * 2 < m) {
    puts("No");
    return 0;
  }
  len = 0;
  for (int i = 1; i <= m; i++) str[++len] = t[i];
  str[++len] = '#';
  for (int i = 1; i <= n; i++) str[++len] = s[i];
  kmp(f);
  len = 0;
  for (int i = 1; i <= n; i++) str[++len] = s[i];
  str[++len] = '#';
  for (int i = 1; i <= m; i++) str[++len] = t[i];
  reverse(str + 1, str + len + 1);
  kmp(g);
  for (int i = m + 2; i <= n + m + 1; i++)
    if (f[i] >= m) {
      int p1 = i - m - 1;
      p1 = max(p1 - 2 * k, 0);
      printf("Yes\n%d %d", p1 + 1, p1 + k + 1);
      return 0;
    }
  for (int i = 0; i <= n; i++) w1[i] = n + 1;
  for (int i = m + 2; i <= n + m + 1; i++)
    if (i - m > k) {
      int tmp = f[i];
      while (tmp) {
        if (i - m <= w1[tmp])
          w1[tmp] = i - m - 1;
        else
          break;
        tmp = f[tmp];
      }
      tmp = g[i];
      while (tmp) {
        if (n - (i - m - 1) >= w2[tmp])
          w2[tmp] = n - (i - m - 1) + 1;
        else
          break;
        tmp = g[tmp];
      }
    }
  w1[0] = 0;
  w2[0] = n + 1;
  for (int i = 1; i <= min(k, m); i++)
    if (i >= m - k) {
      int p1 = w1[i] - k + 1, p2 = w2[m - i];
      if (p1 >= 1 && p2 + k - 1 <= n && w1[i] < p2) {
        printf("Yes\n%d %d", p1, p2);
        return 0;
      }
    }
  puts("No");
}
