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;
vector<int> prefix_function(const string& s, int t) {
  int n = s.length();
  vector<int> pi(n, 0);
  for (int i = 1; i < n; ++i) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) ++j;
    pi[i] = j;
  }
  vector<int> temp(pi.begin() + t, pi.end());
  return move(temp);
}
vector<int> z_function(const string& s, int t) {
  int n = (int)s.length();
  vector<int> z(n);
  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;
  }
  vector<int> temp(z.begin() + t, z.end());
  return move(temp);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  string s, si;
  cin >> n >> m >> k;
  cin >> s >> si;
  string new_str = si + '#' + s;
  vector<int> pf_s = prefix_function(new_str, si.length() + 1);
  reverse(si.begin(), si.end());
  reverse(s.begin(), s.end());
  new_str = si + '#' + s;
  vector<int> pf_b = z_function(new_str, si.length() + 1);
  reverse(pf_b.begin(), pf_b.end());
  vector<set<int>> vals(si.length() + 10);
  for (int i = 0; i < pf_b.size(); ++i) {
    vals[pf_b[i]].insert(i);
  }
  reverse(si.begin(), si.end());
  pair<int, int> answer = make_pair(-1, -1);
  for (int i = k - 1; i < n - k + 1; ++i) {
    int need = m - pf_s[i];
    int res = n;
    if (need > k || pf_s[i] > k) {
      continue;
    }
    auto it = vals[need].upper_bound(i + need - 1);
    if (it != vals[need].end()) {
      res = *it - need + 1;
      if (need == m && res + k - 1 >= n && n - k <= res) {
        res = n - k;
      }
    }
    if (res + k - 1 < n && res > i) {
      answer = make_pair(i - k + 1, res);
    }
  }
  if (answer == make_pair(-1, -1)) {
    for (int i = 0; i < n; ++i) {
      if (pf_s[i] == m && m <= k && i + k < n) {
        int res_l = max(i - k + 1, 0);
        int res_r = res_l + k;
        if (res_r < n) {
          answer = make_pair(res_l, res_r);
          break;
        }
      }
      if (pf_s[i] == m && m > k && m <= 2 * k) {
        int res_l = i - m + 1;
        int res_r = res_l + k;
        if (res_r < n) {
          answer = make_pair(res_l, res_r);
          break;
        }
      }
    }
  }
  if (answer == make_pair(-1, -1)) {
    cout << "No\n";
  } else {
    cout << "Yes\n" << answer.first + 1 << " " << answer.second + 1;
  }
  return 0;
}
