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 = 500050;
const int maxm = 500050;
char s[2][maxn];
int f[maxn];
vector<int> g[maxn];
int beused1[maxn], beused2[maxn];
int n, m, k;
void init() {
  memset(beused1, -1, sizeof(beused1));
  memset(beused2, -1, sizeof(beused2));
}
void getfail(char *s, int len_s) {
  f[0] = f[1] = 0;
  for (int i = 1; i < len_s; i++) {
    int j = f[i];
    while (j && s[i] != s[j]) j = f[j];
    f[i + 1] = (s[i] == s[j]) ? j + 1 : 0;
  }
}
void solve() {
  init();
  scanf("%d %d %d", &n, &m, &k);
  scanf("%s %s", s[0], s[1]);
  if (n < 2 * k || m > 2 * k)
    printf("No\n");
  else {
    getfail(s[1], m);
    int pos0 = 0;
    for (int i = 0; i < n; i++) {
      while (pos0 && s[0][i] != s[1][pos0]) pos0 = f[pos0];
      if (s[0][i] == s[1][pos0]) pos0++;
      if (pos0 == m) {
        printf("Yes\n");
        int l = max(i - 2 * k + 1, 0) + 1, r = l + k;
        printf("%d %d\n", l, r);
        return;
      }
      if (i >= k - 1) {
        int tmp = pos0;
        while (beused1[tmp] == -1) {
          beused1[tmp] = i;
          if (tmp <= k) g[i].push_back(tmp);
          tmp = f[tmp];
        }
      }
    }
    reverse(s[0], s[0] + n);
    reverse(s[1], s[1] + m);
    getfail(s[1], m);
    int pos2 = 0;
    for (int i = 0; i < n; i++) {
      for (auto x : g[n - i - 1]) beused1[x] = -1;
      while (pos2 && s[0][i] != s[1][pos2]) pos2 = f[pos2];
      if (s[0][i] == s[1][pos2]) pos2++;
      if (i >= k - 1) {
        int tmp = pos2;
        while (beused2[tmp] == -1) {
          beused2[tmp] = i;
          if (tmp <= k && m - tmp <= k && beused1[m - tmp] != -1) {
            printf("Yes\n");
            printf("%d %d\n", ((beused1[m - tmp] + 1) - k) + 1,
                   (n - i - 1) + 1);
            return;
          }
          tmp = f[tmp];
        }
      }
    }
    printf("No\n");
  }
}
int main() {
  solve();
  return 0;
}
