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;
using ll = long long;
template <typename t>
void print(ostream& os, const t& a) {
  os << a << '\n';
}
template <typename t, typename... A>
void print(ostream& os, const t& a, A&&... b) {
  os << a << ' ';
  print(os, b...);
}
template <typename t>
using V = vector<t>;
constexpr int oo = 1e9;
int n, m, k;
string sl, wz;
V<int> tb, bt;
void algo(string& sl, string& wz, V<int>& tb, int k) {
  V<int> pref(wz.size());
  ;
  for (int i = 0; i < wz.size(); ++i) {
    int q = (i == 0 ? -1 : pref[i - 1]);
    while (q != -1 and wz[q] != wz[i]) q = (q == 0 ? -1 : pref[q - 1]);
    pref[i] = q + 1;
  };
  tb.resize(wz.size() + 1, oo);
  int cnt = 0;
  for (int i = 0; i < sl.size(); ++i) {
    int q = cnt;
    if (q == wz.size()) q = pref[q - 1];
    while (q != -1 and sl[i] != wz[q]) q = (q == 0 ? -1 : pref[q - 1]);
    ++q;
    cnt = q;
    if (i >= k - 1)
      while (q != -1 and i < tb[q]) tb[q] = i, q = (q == 0 ? -1 : pref[q - 1]);
  };
}
inline int kmp() {
  V<int> pref(wz.size());
  for (int i = 0; i < wz.size(); ++i) {
    int q = (i == 0 ? -1 : pref[i - 1]);
    while (q != -1 and wz[i] != wz[q]) q = (q == 0 ? -1 : pref[q - 1]);
    pref[i] = q + 1;
  }
  int cnt = 0;
  for (int i = 0; i < sl.size(); ++i) {
    int q = cnt;
    if (q == wz.size()) q = pref[wz.size() - 1];
    while (q != -1 and sl[i] != wz[q]) q = (q == 0 ? -1 : pref[q - 1]);
    cnt = q + 1;
    if (q + 1 == wz.size()) return i - wz.size() + 1;
  }
  return oo;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> k;
  cin >> sl >> wz;
  int ind = 0;
  if ((ind = kmp()) != oo) {
    print(cout, "Yes");
    int hi = ind + wz.size();
    int a = hi - k;
    int b = a - k;
    if (b < 0) b = 0, a = k;
    print(cout, b + 1, a + 1);
    return 0;
  }
  algo(sl, wz, tb, k);
  reverse(begin(sl), end(sl));
  reverse(begin(wz), end(wz));
  algo(sl, wz, bt, k);
  for (int i = 0; i < tb.size(); ++i)
    if (tb[i] != oo and bt[wz.size() - i] != oo and
        tb[i] < sl.size() - 1 - bt[wz.size() - i] and i <= k and
        wz.size() - i <= k)
      return print(cout, "Yes"),
             print(cout, tb[i] - k + 2, sl.size() - 1 - bt[wz.size() - i] + 1),
             0;
  print(cout, "No");
}
