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 = 5e5;
char S[maxn + 50], T[maxn + 50];
int len1, len2, k;
int nx[maxn + 50], extend[maxn + 50];
int ans[maxn + 5], ans1[maxn + 5];
int tmp[maxn + 5];
void getnext() {
  nx[0] = len2;
  int j = 0;
  while (j + 1 < len2 && T[j] == T[j + 1]) ++j;
  nx[1] = j;
  int k = 1;
  for (int i = 2; i < len2; ++i) {
    int p = k + nx[k] - 1, l = nx[i - k];
    if (i + l < p + 1)
      nx[i] = l;
    else {
      j = max(p - i + 1, 0);
      while (i + j < len2 && T[i + j] == T[j]) ++j;
      nx[i] = j;
      k = i;
    }
  }
}
void ekmp() {
  int j = 0;
  while (j < len1 && j < len2 && S[j] == T[j]) ++j;
  extend[0] = j;
  int k = 0;
  for (int i = 1; i < len1; ++i) {
    int p = k + extend[k] - 1, l = nx[i - k];
    if (i + l < p + 1)
      extend[i] = l;
    else {
      j = max(p - i + 1, 0);
      while (i + j < len1 && j < len2 && S[i + j] == T[j]) ++j;
      extend[i] = j;
      k = i;
    }
  }
}
set<int> s;
bool cmp(const int a, const int b) { return extend[a] < extend[b]; }
void solve(char *S, char *T, int *ans) {
  getnext();
  ekmp();
  for (int i = 0; i < len1; ++i) tmp[i] = i;
  sort(tmp, tmp + len1, cmp);
  s.clear();
  for (int i = 0; i < len1; ++i) s.insert(i);
  int now = 0;
  for (int x = 0; x <= len2; ++x) {
    while (now < len1 && extend[tmp[now]] < x) s.erase(tmp[now++]);
    set<int>::iterator p = s.lower_bound(k - x);
    if (p == s.end())
      ans[x] = -1;
    else
      ans[x] = *p + x - k;
  }
}
int main() {
  scanf("%d%d%d", &len1, &len2, &k);
  scanf("%s%s", S, T);
  solve(S, T, ans);
  for (int i = 0; i < len1; ++i)
    if (extend[i] == len2) {
      printf("Yes\n");
      int r = len1 - 1;
      int l = r - 2 * k + 1;
      if (l > i) r -= l - i, l -= l - i;
      ++l;
      printf("%d %d\n", l, l + k);
      return 0;
    }
  reverse(S, S + len1);
  reverse(T, T + len2);
  memset(nx, 0, sizeof(nx));
  memset(extend, 0, sizeof(extend));
  solve(S, T, ans1);
  for (int i = 0; i <= len2; ++i)
    if (ans1[i] != -1) ans1[i] = ans1[i] + k - 1;
  for (int i = 0; i <= min(k, len2); ++i)
    if (ans[i] != -1 && ans1[len2 - i] != -1 &&
        ans[i] + k < len1 - ans1[len2 - i] && len2 - i <= k) {
      printf("Yes\n");
      printf("%d %d\n", ans[i] + 1, len1 - ans1[len2 - i]);
      return 0;
    }
  printf("No\n");
  return 0;
}
