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 FN = 500005;
long long ft[2][FN];
void chw(int i, int vl, bool b) {
  if (i == 0) return;
  while (i < FN) {
    ft[b][i] += vl;
    i += (i & -i);
  }
}
void psh(int i, int vl) {
  chw(1, vl, 0);
  chw(i + 1, -vl, 0);
  chw(i + 1, vl * i, 1);
}
void shs(int l, int r, int vl) {
  psh(r, vl);
  psh(l - 1, -vl);
}
int pj(int i, bool b) {
  int ds = 0;
  while (i) {
    ds += ft[b][i];
    i -= (i & -i);
  }
  return ds;
}
int pjs(int i) { return pj(i, 0) * i + pj(i, 1); }
int sj(int l, int r) { return pjs(r) - pjs(l - 1); }
void clr() {
  for (int i = 0; i < FN; ++i) {
    ft[0][i] = ft[1][i] = 0;
  }
}
int n, m, k;
string s, t;
vector<int> a, b, c;
vector<int> dkf(string a, string b) {
  a = b + a;
  vector<int> dr, rd;
  dr.resize(a.size(), 0);
  int l = 0, r = 0;
  for (int i = 1; i < a.size(); ++i) {
    if (r >= i) dr[i] = min(r - i + 1, dr[i - l]);
    while (dr[i] + i < a.size() && a[dr[i]] == a[dr[i] + i]) ++dr[i];
    if (dr[i] + i - 1 > r) r = dr[i] + i - 1, l = i;
    if (i >= b.size()) rd.push_back(min(dr[i], (int)b.size()));
  }
  return rd;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k >> s >> t;
  a = dkf(s, t);
  reverse(s.begin(), s.end());
  reverse(t.begin(), t.end());
  b = dkf(s, t);
  reverse(b.begin(), b.end());
  c.resize(n, 0);
  for (int i = 0; i < n; ++i) {
    a[i] = min(a[i], k);
    b[i] = min(b[i], k);
    if (a[i] == m) {
      cout << "Yes\n";
      if (i + 2 * k - 1 < n)
        cout << i + 1 << " " << i + 1 + k << endl;
      else
        cout << n - 2 * k << " " << n - k << endl;
      return 0;
    }
    if (i < k)
      c[i] = k - i;
    else
      c[i] = 1;
    if (i - m + 1 >= 0) {
      if (c[i - m + 1] <= a[i - m + 1]) {
        shs(c[i - m + 1], a[i - m + 1], 1);
      }
    }
    int l;
    if (i < n - k)
      l = 1;
    else
      l = i - n + k + 1;
    if (l <= b[i]) {
      if (sj(m - b[i], m - l)) {
        cout << "Yes\n";
        for (int j = 0; j < n; ++j) {
          if (max(m - b[i], c[j]) <= min(m - l, a[j])) {
            if (a[j] >= m - b[i] && a[j] <= m - l) {
              cout << j - k + a[j] + 1 << " " << i - m + a[j] + 2 << endl;
              return 0;
            }
            if (c[j] >= m - b[i] && c[j] <= m - l) {
              cout << j - k + c[j] + 1 << " " << i - m + c[j] + 2 << endl;
              return 0;
            }
            if (m - b[i] >= c[j] && m - b[i] <= a[j]) {
              cout << j - k + m - b[i] + 1 << " " << i - b[i] + 2 << endl;
              return 0;
            }
            if (m - l >= c[j] && m - l <= a[j]) {
              cout << j - k + m - l + 1 << " " << i - l + 2 << endl;
              return 0;
            }
          }
        }
      }
    }
  }
  cout << "No" << endl;
  return 0;
}
