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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << endl;
}
void getZarr(string str, vector<int>& Z);
void func(string str, vector<int>& vec) {
  int l = str.length();
  vec.resize(l);
  getZarr(str, vec);
}
void getZarr(string str, vector<int>& Z) {
  int n = str.length();
  int L, R, k;
  L = R = 0;
  for (int i = 1; i < n; ++i) {
    if (i > R) {
      L = R = i;
      while (R < n && str[R - L] == str[R]) R++;
      Z[i] = R - L;
      R--;
    } else {
      k = i - L;
      if (Z[k] < R - i + 1)
        Z[i] = Z[k];
      else {
        L = i;
        while (R < n && str[R - L] == str[R]) R++;
        Z[i] = R - L;
        R--;
      }
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  int ttt = 1;
  while (ttt--) {
    int n, m, k;
    int i;
    cin >> n >> m >> k;
    string str1, str2;
    cin >> str1 >> str2;
    string temp = str2 + "$" + str1;
    vector<int> vec1;
    func(temp, vec1);
    for (i = m + 1; i < vec1.size(); i++) {
      if (vec1[i] == m) {
        cout << "Yes" << endl;
        int idx1 = i - m;
        int idx2 = idx1 + m - 1;
        int l = idx2 + 1 - 2 * k;
        if (l >= 1) {
          cout << l << " " << l + k;
        } else {
          cout << "1 " << k + 1;
        }
        return 0;
      }
    }
    set<int> set1;
    for (i = 1; i <= m; i++) {
      set1.insert(i);
    }
    vector<int> left(m + 1, -1);
    for (i = m + 1; i < vec1.size(); i++) {
      int idx = i - m;
      set<int>::iterator it1;
      while (1) {
        int minn = k - idx + 1;
        int maxx = vec1[i];
        it1 = set1.upper_bound(min(k, vec1[i]));
        if (it1 == set1.begin()) break;
        it1--;
        if (*it1 < minn) break;
        left[*it1] = idx;
        set1.erase(it1);
      }
    }
    reverse(str2.begin(), str2.end());
    reverse(str1.begin(), str1.end());
    temp = str2 + "$" + str1;
    vector<int> vec2;
    func(temp, vec2);
    set1.clear();
    for (i = 1; i <= m; i++) {
      set1.insert(i);
    }
    vector<int> right(m + 1, -1);
    for (i = m + 1; i < vec2.size(); i++) {
      int idx = i - m;
      idx = n - idx + 1;
      set<int>::iterator it1;
      while (1) {
        int maxx = vec2[i];
        it1 = set1.upper_bound(min(k, vec2[i]));
        if (it1 == set1.begin()) break;
        it1--;
        int x = *it1;
        int y = idx - x + 1;
        int z = y + k - 1;
        if (z > n) break;
        right[*it1] = idx;
        set1.erase(it1);
      }
    }
    for (i = 1; i < m; i++) {
      int a = left[i];
      int b = right[m - i];
      if (a == -1 || b == -1) continue;
      int a2 = left[i] + i - 1;
      int b2 = b - (m - i) + 1;
      if (a2 >= b2) continue;
      cout << "Yes" << endl;
      cout << a2 - k + 1 << " " << b2 << endl;
      return 0;
    }
    cout << "No";
  }
  return 0;
}
