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 inf = 0x3f3f3f3f;
const int N = 500000, M = 500000;
int n, m, s, t;
int rev_pos(int pos) { return n + 1 - pos; }
char a[N + 5], b[M + 5], ra[N + 5], rb[M + 5], c[N + 1 + M + 5];
void con(char str1[], char str2[]) {
  t = 0;
  for (int i = 1; i <= m; i++) c[++t] = str1[i];
  c[++t] = '!';
  for (int i = 1; i <= n; i++) c[++t] = str2[i];
}
int z1[N + 1 + M + 1], z2[N + 1 + M + 1];
void z_init(int z[]) {
  int zl = 0, zr = 0;
  for (int i = 2; i <= t; i++)
    if (zr < i) {
      while (i + z[i] <= t && c[i + z[i]] == c[1 + z[i]]) z[i]++;
      if (z[i]) zl = i, zr = i + z[i] - 1;
    } else if (i + z[i - zl + 1] <= zr)
      z[i] = z[i - zl + 1];
    else {
      z[i] = zr - i + 1;
      while (i + z[i] <= t && c[i + z[i]] == c[1 + z[i]]) z[i]++;
      zl = i;
      zr = i + z[i] - 1;
    }
}
int lft[M + 1], rit[M + 1];
vector<int> dadd[M + 1], ddel[N + 1];
multiset<int> st;
int main() {
  cin >> n >> m >> s >> a + 1 >> b + 1;
  memcpy(ra + 1, a + 1, n + 1);
  reverse(ra + 1, ra + n + 1);
  memcpy(rb + 1, b + 1, m + 1);
  reverse(rb + 1, rb + m + 1);
  con(b, a);
  z_init(z1);
  con(rb, ra);
  z_init(z2);
  if (s >= m)
    for (int i = 1; i <= n; i++)
      if (z1[m + 1 + i] == m) {
        int l = max(1, i - (s - m)), r = l + s;
        if (r + s - 1 <= n) return cout << "Yes\n" << l << " " << r, 0;
        r = min(n, i + s - 1) - s + 1;
        l = r - s;
        if (l >= 1) return cout << "Yes\n" << l << " " << r, 0;
      }
  for (int i = 1; i <= n; i++) {
    int l = max(s - (i - 1), 1), r = z1[m + 1 + i];
    if (l > r) continue;
    dadd[l].push_back(i - s);
    if (r < m) ddel[r + 1].push_back(i - s);
  }
  st.insert(inf);
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < dadd[i].size(); j++) st.insert(dadd[i][j]);
    for (int j = 0; j < ddel[i].size(); j++) st.erase(ddel[i][j]);
    lft[i] = *st.begin() + i;
  }
  for (int i = 1; i <= m; i++) dadd[i].clear(), ddel[i].clear();
  for (int i = 1; i <= n; i++) {
    int l = max(s - (n - i), 1), r = z2[m + 1 + rev_pos(i)];
    if (l > r) continue;
    dadd[l].push_back(i + 1);
    if (r < m) ddel[r + 1].push_back(i + 1);
  }
  st.clear();
  st.insert(-inf);
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < dadd[i].size(); j++) st.insert(dadd[i][j]);
    for (int j = 0; j < ddel[i].size(); j++) st.erase(ddel[i][j]);
    rit[i] = *--st.end() - i;
  }
  for (int i = 0; i <= s; i++)
    if (0 <= m - i && m - i <= s)
      if (lft[i] + s - 1 < rit[m - i])
        return cout << "Yes\n" << lft[i] << " " << rit[m - i], 0;
  puts("No");
  return 0;
}
