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;
int F[500010];
int fwd[500010], bkd[500010], K, r = 0;
void build_failure_function(string &pattern) {
  int m = pattern.length();
  F[0] = F[1] = 0;
  F[0] = -1;
  for (int i = 2; i <= m; i++) {
    int j = F[i - 1];
    while (1) {
      if (pattern[j] == pattern[i - 1]) {
        F[i] = j + 1;
        break;
      }
      if (j == 0) {
        F[i] = 0;
        break;
      }
      j = F[j];
    }
  }
  return;
}
void KMP(string &text, string &pattern, int flag = 0) {
  int n = text.length(), m = pattern.length();
  build_failure_function(pattern);
  int i = 0, j = 0;
  while (1) {
    if (j == n) break;
    if (text[j] == pattern[i]) {
      int k = i;
      while (k != -1) {
        if (!flag) {
          if (fwd[k] == -1 && (j >= K - 1 || r))
            fwd[k] = j;
          else
            break;
        } else {
          if (bkd[k] == -1 && (j >= K - 1 || r))
            bkd[k] = j;
          else
            break;
        }
        k = F[k + 1] - 1;
      }
      i++;
      j++;
      if (i == m) return;
    } else if (i > 0)
      i = F[i];
    else
      j++;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  if (m < k) r = 1;
  K = k;
  string s, t;
  cin >> s >> t;
  for (int i = 0; i < m; i++) fwd[i] = bkd[i] = -1;
  KMP(s, t);
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  KMP(s, t, 1);
  for (int i = 0; i < m; i++)
    if (bkd[i] != -1) bkd[i] = n - 1 - bkd[i];
  for (int i = 0; i < m / 2; i++) swap(bkd[i], bkd[m - 1 - i]);
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  bkd[m] = 0;
  for (int i = 0; i < m; i++) {
    if (fwd[i] == -1 || bkd[i + 1] == -1) continue;
    if (i == m - 1) {
      int start = max(0, fwd[i] - 2 * k + 1);
      cout << "Yes\n";
      cout << start + 1 << " " << start + k + 1 << "\n";
      return 0;
    }
    if (i + 1 > k || (m - 1 - i) > k) continue;
    int start = fwd[i] - k + 1;
    if (start < 0) continue;
    if (bkd[i + 1] > fwd[i] && bkd[i + 1] + k - 1 < n) {
      cout << "Yes\n";
      cout << start + 1 << " " << bkd[i + 1] + 1 << "\n";
      return 0;
    }
  }
  cout << "No\n";
  return 0;
}
