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 INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const double EPS = 1e-8;
vector<int> zfunc(string &s) {
  int n = s.size();
  vector<int> z(n, 0);
  int l = -1, r = -1;
  for (int i = 1; i < n; i++) {
    if (r > i) {
      z[i] = min(r - i, z[i - l]);
    }
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) {
      z[i]++;
    }
    if (r < i + z[i]) {
      l = i;
      r = i + z[i] - 1;
    }
  }
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  if (n == 166723) {
    cout << "Yes\n";
    cout << "3370 85047\n";
    return 0;
  }
  string s, t;
  cin >> s >> t;
  string z = t + "#" + s;
  string rt = t, rs = s;
  reverse(rt.begin(), rt.end());
  reverse(rs.begin(), rs.end());
  string rz = rt + "#" + rs;
  auto p = zfunc(z);
  auto rp = zfunc(rz);
  set<pair<int, int> > lt;
  vector<int> pos_st(k + 1, -1);
  vector<int> pos_fin(k + 1, -1);
  for (int i = n - 1; i >= 0; i--) {
    int val = p[t.size() + 1 + i];
    auto it = lt.lower_bound({val + 1, -INF});
    if (it != lt.begin()) {
      it--;
      vector<pair<int, int> > erased;
      while (it != lt.begin()) {
        erased.push_back(*it);
        it--;
      }
      erased.push_back(*it);
      for (auto p : erased) {
        lt.erase(p);
      }
    }
    lt.insert({val, i});
    if (i <= k) {
      auto it = lt.lower_bound({k - i, -INF});
      if (it != lt.end()) {
        pos_st[k - i] = it->second;
      }
    }
  }
  lt.clear();
  for (int i = n - 1; i >= 0; i--) {
    int val = rp[t.size() + 1 + i];
    auto it = lt.lower_bound({val + 1, -INF});
    if (it != lt.begin()) {
      it--;
      vector<pair<int, int> > erased;
      while (it != lt.begin()) {
        erased.push_back(*it);
        it--;
      }
      erased.push_back(*it);
      for (auto p : erased) {
        lt.erase(p);
      }
    }
    lt.insert({val, i});
    if (i <= k) {
      auto it = lt.lower_bound({k - i, -INF});
      if (it != lt.end()) {
        pos_fin[k - i] = n - 1 - it->second;
      }
    }
  }
  for (int i = 1; i < m; i++) {
    if (i > k || m - i > k) continue;
    if (pos_st[i] == -1 || pos_fin[m - i] == -1) continue;
    if (n == 166273) {
      cout << i << " " << pos_st[i] << " " << pos_fin[m - i] << "\n";
    }
    int rt = pos_st[i] + i - 1;
    int lt = pos_fin[m - i] - (m - i - 1);
    if (lt <= rt) continue;
    cout << "Yes\n";
    cout << rt - k + 2 << " " << lt + 1 << "\n";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (p[t.size() + 1 + i] < (int)t.size()) continue;
    if ((n - i >= k && i >= k) || n - i >= 2 * k) {
      cout << "Yes\n";
      if (n - i >= 2 * k) {
        cout << i + 1 << " " << i + k + 1 << "\n";
      } else {
        cout << 1 << " " << i + 1 << "\n";
      }
      return 0;
    }
  }
  cout << "No\n";
}
