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 = 1000005;
const int Inf = 1000000000;
int n, m, k;
char S[Maxn];
char T[Maxn];
char str[Maxn];
int slen;
int Z[Maxn];
int res1[Maxn];
int res2[Maxn];
set<int> St;
void Solve(int res[]) {
  St.clear();
  for (int i = 0; i <= m; i++) St.insert(i);
  fill(res, res + m + 1, Inf);
  int L = -1, R = -1;
  for (int i = 1; i < slen; i++) {
    if (i <= R)
      Z[i] = min(R - i, Z[i - L]);
    else
      Z[i] = 0;
    while (i + Z[i] < slen && str[i + Z[i]] == str[Z[i]]) Z[i]++;
    if (i >= m + 1) {
      int tk = min(Z[i], k);
      int ind = i - m - 1;
      if (tk + ind >= k) {
        int lef = max(0, k - ind);
        int rig = tk;
        if (lef <= rig) {
          auto it = St.lower_bound(lef);
          while (it != St.end() && *it <= rig) {
            res[*it] = ind;
            St.erase(it++);
          }
        }
      } else if (tk == m) {
        int mnind = max(0, ind + tk - k);
        auto it = St.lower_bound(m);
        if (it != St.end()) {
          res[*it] = mnind;
          St.erase(it);
        }
      }
    }
    if (i + Z[i] - 1 > R) {
      R = i + Z[i] - 1;
      L = i;
    }
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  scanf("%s", S);
  scanf("%s", T);
  for (int i = 0; i < m; i++) str[slen++] = T[i];
  str[slen++] = '#';
  for (int i = 0; i < n; i++) str[slen++] = S[i];
  Solve(res1);
  slen = 0;
  for (int i = m - 1; i >= 0; i--) str[slen++] = T[i];
  str[slen++] = '#';
  for (int i = n - 1; i >= 0; i--) str[slen++] = S[i];
  Solve(res2);
  for (int i = 0; i <= m; i++) {
    int j = m - i;
    if (res1[i] + i + res2[j] + j <= n) {
      printf("Yes\n");
      printf("%d %d\n", max(1, res1[i] + i - k + 1),
             min(n - k + 1, n - res2[j] - j + 1));
      return 0;
    }
  }
  printf("No\n");
  return 0;
}
