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 n, m, k;
string s1, t1;
int z[2][1000005];
int ll[1000005], rl[1000005];
int lm[1000005];
void Z(string s, int p) {
  int l = 0, r = 0;
  int ls = s.size();
  for (int i = 1; i < ls; i++) {
    if (i <= r) z[p][i] = min(z[p][i - l], r - i);
    while (i + z[p][i] < ls && s[z[p][i]] == s[i + z[p][i]]) z[p][i]++;
    if (i + z[p][i] > r) {
      l = i;
      r = i + z[p][i];
    }
  }
}
int main() {
  cin >> n >> m >> k;
  cin >> s1 >> t1;
  string s2 = s1;
  reverse(s2.begin(), s2.end());
  string t2 = t1;
  reverse(t2.begin(), t2.end());
  string S1 = t1 + "#" + s1;
  string S2 = t2 + "#" + s2;
  Z(S1, 0);
  Z(S2, 1);
  for (int i = 0; i + k <= n; i++) {
    rl[i] = min(k, z[0][m + 1 + i]);
    if (rl[i] >= m && i + rl[i] < k && k < n) {
      cout << "Yes" << endl;
      cout << 1 << " " << 1 + k;
      return 0;
    }
  }
  for (int i = k - 1; i < n; i++) {
    ll[i] = min(k, z[1][m + n - i]);
    if (ll[i] >= m && i - ll[i] >= n - k) {
      cout << "Yes" << endl;
      cout << 1 << " " << n - k + 1;
      return 0;
    }
  }
  for (int i = 0; i + k <= n; i++)
    if (i + rl[i] - k >= 0) lm[i] = i;
  for (int i = 1; i < n; i++)
    if (rl[lm[i - 1]] > rl[lm[i]]) lm[i] = lm[i - 1];
  for (int i = max(m - 1, k - 1); i < n; i++) {
    int j = lm[i - m + 1];
    if (rl[j] + ll[i] >= m && j + rl[j] - k >= 0 && i - m + rl[j] + 1 >= 0 &&
        i - m + rl[j] + k < n) {
      cout << "Yes" << endl;
      cout << j + rl[j] - k + 1 << " " << i - m + rl[j] + 2;
      return 0;
    }
  }
  cout << "No";
  return 0;
}
