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 MAXN = 500005;
int n, m, k;
string s, t;
int back[MAXN], dp[MAXN][2], pos[MAXN];
void build() {
  int curr = -1;
  back[0] = -1;
  for (int i = 1; i < m; i++) {
    while (curr != -1 && t[i] != t[curr + 1]) {
      curr = back[curr];
    }
    curr++;
    back[i] = curr;
    if (curr == 0 && t[i] != t[0]) back[i] = -1, curr--;
  }
}
void calc_dp(int ind) {
  int curr = 0;
  dp[0][ind] = 0;
  if (s[0] != t[0]) dp[0][ind] = -1, curr--;
  for (int i = 1; i < n; i++) {
    while (curr != -1 && s[i] != t[curr + 1]) {
      curr = back[curr];
    }
    curr++;
    dp[i][ind] = curr;
    if (curr == 0 && s[i] != t[0]) dp[i][ind] = -1, curr--;
  }
}
void nadi(int x, int y) {
  for (int i = 0; i < n; i++) {
    if (i <= x && y <= i + 2 * k - 1) {
      cout << "Yes" << endl << i + 1 << " " << i + 1 + k;
      break;
    }
  }
}
int main() {
  memset(pos, -1, sizeof pos);
  cin >> n >> m >> k >> s >> t;
  build();
  calc_dp(0);
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  build();
  calc_dp(1);
  reverse(t.begin(), t.end());
  build();
  for (int i = 0; i < n; i++) {
    if (dp[i][0] == m - 1) {
      nadi(i - m + 1, i);
      return 0;
    }
    if (dp[i][1] == m - 1) {
      nadi(i, i + m - 1);
      return 0;
    }
  }
  for (int i = k - 1; i < n - k; i++) {
    int val = dp[i][0], q = dp[n - i - 2][1];
    while (val < k && val != -1 && pos[val] == -1) {
      pos[val] = i;
      val = back[val];
    }
    if (q >= k || q == -1) continue;
    if (m - q - 2 >= 0 && pos[m - q - 2] != -1) {
      cout << "Yes" << endl << pos[m - q - 2] - k + 2 << " " << i + 2;
      return 0;
    }
  }
  cout << "No";
  return 0;
}
