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>
const int maxn = 1000001;
const double EPS = 1e-9;
const double PI = 3.141592653589793;
using namespace std;
const int MOD = 1e9 + 7;
vector<int> z_func(string& s) {
  int n = s.size();
  vector<int> z(n);
  assert(!z[0]);
  int l = -1, r = -1;
  for (int i = 1; i < n; ++i) {
    if (i <= r) {
      z[i] = min(z[i - l], r - i + 1);
    }
    while (i + z[i] < n && s[i + z[i]] == s[z[i]]) ++z[i];
    if (i + z[i] - 1 > r) {
      r = i + z[i] - 1;
      l = i;
    }
  }
  return z;
}
int pref[maxn];
int suff[maxn];
vector<int> g[maxn];
vector<int> gr[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  string s;
  cin >> s;
  string t;
  cin >> t;
  string temp = t + "#" + s;
  auto z_pref = z_func(temp);
  string ss = s;
  reverse(ss.begin(), ss.end());
  string tt = t;
  reverse(tt.begin(), tt.end());
  string temp2 = tt + "#" + ss;
  auto z_suff = z_func(temp2);
  set<int> se;
  for (int i = m + 1; i < temp.size(); ++i) {
    if (z_pref[i] >= k) se.insert(i - m - 1);
    g[z_pref[i]].push_back(i - m - 1);
  }
  for (int i = k; i >= 1; --i) {
    for (int el : g[i]) se.insert(el);
    auto it = se.upper_bound(k - 1 - i);
    if (it == se.end())
      pref[i] = -1;
    else
      pref[i] = *it;
  }
  vector<int> r_suff(n);
  for (int i = 0; i < n; ++i) {
    r_suff[i] = z_suff[i + m + 1];
  }
  for (int i = 0; i < n; ++i) {
    if (i < n - i - 1) swap(r_suff[i], r_suff[n - i - 1]);
  }
  se.clear();
  for (int i = 0; i < n; ++i) {
    gr[r_suff[i]].push_back(i);
    if (r_suff[i] >= k) se.insert(i);
  }
  for (int i = k; i >= 1; --i) {
    for (int el : gr[i]) se.insert(el);
    auto it = se.lower_bound(n - k + i);
    if (it == se.begin()) {
      suff[i] = -1;
    } else {
      --it;
      suff[i] = *it;
    }
  }
  for (int i = m + 1; i < temp.size(); ++i) {
    if (z_pref[i] >= m) {
      int ind = min(i + 2 * k - 1, n + m);
      cout << "Yes\n";
      cout << ind - 2 * k + 1 - m << " " << ind - k + 1 - m << "\n";
      return 0;
    }
  }
  for (int i = 1; i <= k && i < m; ++i) {
    if (suff[m - i] != -1 && pref[i] != -1 && suff[m - i] - pref[i] + 1 >= m) {
      cout << "Yes\n";
      cout << pref[i] - (k - i) + 1 << " " << suff[m - i] - (m - i) + 2 << "\n";
      return 0;
    }
  }
  cout << "No";
}
