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;
using ld = long double;
using pll = pair<ll, ll>;
using vll = vector<ll>;
using vpll = vector<pll>;
using gr = vector<vll>;
using wgr = vector<vpll>;
struct d_ {
  template <typename T>
  d_& operator,(const T& x) {
    cerr << ' ' << x;
    return *this;
  }
  template <typename T>
  d_& operator,(const vector<T>& x) {
    for (auto x : x) cerr << ' ' << x;
    return *this;
  }
} d_t;
ll B[500007];
void buildTable(string& patt) {
  (memset(B, 0, sizeof(B)));
  B[0] = -1;
  for (ll i(0); i < patt.size(); i++) {
    int j = B[i];
    while (j != -1 && patt[j] != patt[i]) j = B[j];
    B[i + 1] = j + 1;
  }
}
vll kmp(string& txt, string& patt, unordered_map<int, int>& occ, int start) {
  vll foundpos;
  vll res(txt.size(), -1);
  buildTable(patt);
  ll pos = 0;
  for (int i = 0; i < txt.size(); ++i) {
    while (pos != -1 && patt[pos] != txt[i]) pos = B[pos];
    ll ns = patt.size();
    pos++;
    if (pos == ns) {
      res[i] = pos;
      int t = pos;
      while (i >= start && occ.find(t) == occ.end()) {
        occ[t] = i;
        t = B[t];
      }
      pos = B[ns];
    } else {
      res[i] = pos;
      int t = pos;
      while (i >= start && occ.find(t) == occ.end()) {
        occ[t] = i;
        t = B[t];
      }
    }
  }
  return res;
}
int main() {
  ll n, m, k;
  cin >> n >> m >> k;
  string s, t;
  cin >> s >> t;
  string prefT = t.substr(0, k);
  unordered_map<int, int> prefs;
  vll temp = kmp(s, prefT, prefs, k - 1);
  for (ll i(0); i < temp.size(); i++) {
    vector<int> res(2);
    if (temp[i] == m) {
      int start = max(0LL, i - 2 * k + 1);
      cout << "Yes" << endl;
      cout << start + 1 << " " << start + k + 1 << endl;
      return 0;
    }
  }
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  string suffT = t.substr(0, k);
  unordered_map<int, int> sufs;
  temp = kmp(s, suffT, sufs, k - 1);
  reverse(temp.begin(), temp.end());
  sufs[0] = k - 1;
  prefs[0] = k - 1;
  for (int i = 0; i <= k; ++i) {
    if (prefs.find(i) != prefs.end() && sufs.find(m - i) != sufs.end()) {
      if (prefs[i] < (n - 1 - sufs[m - i])) {
        cout << "Yes" << endl;
        cout << prefs[i] - k + 2 << " " << (n - 1 - sufs[m - i]) + 1 << endl;
        return 0;
      }
    }
  }
  cout << "No" << endl;
  return 0;
}
