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 mn = 5e5 + 5;
int n, m, k, j, t, bo, nxt[mn], ans[mn];
char a[mn], b[mn];
inline void swap(char &a, char &b) {
  char c = a;
  a = b, b = c;
}
inline int Min(int a, int b) { return a < b ? a : b; }
inline int Max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d%d%d\n%s\n%s", &n, &m, &k, a + 1, b + 1);
  for (register int i = 2; i <= m; ++i) {
    while (j && b[i] != b[j + 1]) j = nxt[j];
    if (b[i] == b[j + 1]) ++j;
    nxt[i] = j;
  }
  j = 0;
  for (register int i = 1; i <= n; ++i) {
    while (j && a[i] != b[j + 1]) j = nxt[j];
    if (a[i] == b[j + 1]) ++j;
    t = Min(i - m + 1, n - k + 1);
    if (j == m && t - 1 >= k) printf("Yes\n%d %d", 1, t), exit(0);
    t = Max(i - Max(k, m) + 1, 1);
    if (j == m && t + k * 2 - 1 <= n) printf("Yes\n%d %d", t, t + k), exit(0);
    !ans[j] && j <= k &&i >= k && (ans[j] = i);
  }
  for (int i = k; i; --i)
    nxt[i] && ans[i] && (ans[nxt[i]] = Min(ans[nxt[i]], ans[i]));
  t = n >> 1;
  for (register int i = 1; i <= t; ++i) swap(a[i], a[n - i + 1]);
  t = m >> 1;
  for (register int i = 1; i <= t; ++i) swap(b[i], b[m - i + 1]);
  j = 0;
  for (register int i = 2; i <= m; ++i) {
    while (j && b[i] != b[j + 1]) j = nxt[j];
    if (b[i] == b[j + 1]) ++j;
    nxt[i] = j;
  }
  j = 0;
  for (register int i = 1; i <= n; ++i) {
    while (j && a[i] != b[j + 1]) j = nxt[j];
    if (a[i] == b[j + 1]) ++j;
    if (i >= k && j <= k && ans[m - j] >= k && ans[m - j] + i <= n) {
      bo = 1, printf("Yes\n%d %d", ans[m - j] - k + 1, n - i + 1);
      break;
    }
  }
  return !bo && printf("No"), 0;
}
