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;
char pat[500004], txt[500004];
int N, M, k;
int dp[500004], dp2[500004];
int lps[500004];
int ans1 = -1, ans2 = -1;
void KMPSearch(bool flag) {
  int i = 0;
  int j = 0;
  while (i < N) {
    if (pat[j] == txt[i]) {
      int temp = j;
      if (i + 1 >= k && dp[j + 1] == 0 && flag) {
        dp[j + 1] = i + 1;
      }
      if (i + 1 >= k && dp2[j + 1] == 0 && !flag) {
        dp2[j + 1] = i + 1;
      }
      j++;
      i++;
    }
    if (j == M) {
      if (flag && i < k) {
        ans1 = 1;
        ans2 = k + 1;
        return;
      }
      if (!flag && i < k) {
        ans1 = 1;
        ans2 = N - k + 1;
        return;
      }
      if (flag && k < i - k + 1 && M <= k) {
        ans1 = 1;
        ans2 = i - k + 1;
        return;
      }
      if (flag && i >= k && N - k + 1 > i && M <= k) {
        ans1 = i - k + 1;
        ans2 = N - k + 1;
        return;
      }
      if (flag && i >= 2 * k && M <= k) {
        ans1 = 1;
        ans2 = i - k + 1;
        return;
      }
      j = max(0, lps[j - 1]);
    } else if (i < N && pat[j] != txt[i]) {
      if (j > 0) {
        j = max(0, lps[j - 1]);
      } else {
        i = i + 1;
      }
    }
  }
}
void computeLPSArray() {
  int len = 0;
  lps[0] = -1;
  int i = 1;
  while (i < M) {
    if (pat[i] == pat[len]) {
      len++;
      lps[i] = len;
      i++;
    } else {
      if (len >= 1) {
        len = max(0, lps[len - 1]);
      } else {
        lps[i] = -1;
        len = 0;
        i++;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int i, j, l;
  cin >> N >> M >> k;
  cin >> txt >> pat;
  computeLPSArray();
  KMPSearch(1);
  if (ans1 != -1) {
    cout << "Yes\n" << ans1 << " " << ans2;
    return 0;
  }
  for (i = M - 1; i >= 0; i--) {
    if (!dp[i + 1]) continue;
    j = lps[i];
    if (j != -1) {
      if (dp[j] == 0)
        dp[j] = dp[i + 1];
      else
        dp[j] = min(dp[j], dp[i + 1]);
    }
  }
  for (i = 0; i <= ((M - 1) / 2); i++) {
    j = M - 1 - i;
    char temp = pat[i];
    pat[i] = pat[j];
    pat[j] = temp;
  }
  for (i = 0; i <= ((N - 1) / 2); i++) {
    j = N - 1 - i;
    char temp = txt[i];
    txt[i] = txt[j];
    txt[j] = temp;
  }
  computeLPSArray();
  KMPSearch(0);
  if (ans1 != -1) {
    cout << "Yes\n" << ans1 << " " << ans2;
    return 0;
  }
  for (i = M - 1; i >= 0; i--) {
    if (!dp2[i + 1]) continue;
    j = lps[i];
    if (j != -1) {
      if (dp2[j] == 0)
        dp2[j] = dp2[i + 1];
      else
        dp2[j] = min(dp2[j], dp2[i + 1]);
    }
  }
  dp[0] = k;
  dp2[0] = k;
  for (i = 0; i <= M; i++) {
    if (dp[i] && dp2[M - i] && dp[i] < N - dp2[M - i] + 1 && i <= k &&
        (M - i) <= k) {
      cout << "Yes\n" << dp[i] - k + 1 << " " << N - dp2[M - i] + 1;
      return 0;
    }
  }
  cout << "No";
  return 0;
}
