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;
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
vector<int> z_function(string s) {
  int n = (int)s.length();
  vector<int> z(n, 0);
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  string s, t;
  cin >> s >> t;
  string t_s(t);
  t_s.push_back('#');
  t_s += s;
  vector<int> z_t_s = z_function(t_s);
  string inv_s_t(s);
  inv_s_t.push_back('#');
  inv_s_t += t;
  reverse(inv_s_t.begin(), inv_s_t.end());
  vector<int> z_inv_s_t = z_function(inv_s_t);
  multimap<int, int> rev_map;
  for (int j = m - 1; j < n; ++j) {
    rev_map.insert(make_pair(min(z_inv_s_t[m + n - j], k), j));
  }
  for (int i = 0; i < n - m + 1; ++i) {
    int m1 = min(k, z_t_s[m + 1 + i]);
    int m2 = m - m1;
    if (m2 == 0) {
      cout << "Yes" << endl;
      int l = max(i + m / 2 - k, 0);
      int r = l + k;
      if (r + k > n) {
        r = n - k;
        l = r - k;
      }
      cout << l + 1 << " " << r + 1;
      return 0;
    }
    if (i + m1 - k < 0) continue;
    if (rev_map.upper_bound(m2 - 1) == rev_map.end()) continue;
    for (auto j_it = rev_map.upper_bound(m2 - 1); j_it != rev_map.end();
         ++j_it) {
      int j = j_it->second;
      if (m1 == 0) {
        cout << "Yes" << endl;
        int l = max(j - m2 + 1 + m / 2 - k, 0);
        int r = l + k;
        if (r + k > n) {
          r = n - k;
          l = r - k;
        }
        cout << l + 1 << " " << r + 1;
        return 0;
      } else if (j - m2 + k < n && j - m2 >= i + m1 - 1) {
        cout << "Yes" << endl;
        int l = i + m1 - k;
        int r = j - m2 + 1;
        cout << l + 1 << " " << r + 1;
        return 0;
      }
    }
  }
  cout << "No";
  return 0;
}
