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 long long INF = 0x3f3f3f3f3f3f3f3fll;
const int maxn = 500005 << 1;
char S[maxn], T[maxn];
int pi[maxn];
void init() {
  int k = -1;
  pi[0] = -1;
  for (int i = 1; T[i]; i++) {
    while (k != -1 && T[k + 1] != T[i]) k = pi[k];
    pi[i] = ++k;
  }
}
int lpos[maxn], rpos[maxn];
int K;
int tmpsubmit;
void solve(char *str, int *pos) {
  memset(pos, 0x3f, sizeof(int) * maxn);
  int k = 0;
  int lenT = strlen(T + 1);
  for (int i = 1; str[i]; i++) {
    while (k != -1 && str[i] != T[k + 1]) k = pi[k];
    int len = ++k;
    len = min(len, K);
    if (len == lenT) {
      int st1 = max(i - K + 1, 1);
      int st2 = st1 + K;
      if (str[st2 + K - 1]) {
        puts("Yes");
        printf("%d %d\n", st1, st2);
        exit(0);
      }
      k = pi[k];
    }
    if (i >= K) {
      pos[len] = min(pos[len], i - K + 1);
    }
  }
  int len = strlen(str + 1);
  for (int i = len; i >= 1; i--) pos[pi[i]] = min(pos[i], pos[pi[i]]);
  pos[0] = 1;
}
int main() {
  int n, m;
  scanf("%d%d%d", &n, &m, &K);
  scanf("%s", S + 1);
  scanf("%s", T + 1);
  init();
  solve(S, lpos);
  reverse(S + 1, S + 1 + strlen(S + 1));
  reverse(T + 1, T + 1 + strlen(T + 1));
  init();
  solve(S, rpos);
  int len = strlen(T + 1);
  int lenS = strlen(S + 1);
  for (int i = 0; i <= len; i++) {
    int res = len - i;
    if (lpos[i] == 0x3f3f3f3f || rpos[res] == 0x3f3f3f3f) continue;
    if (lpos[i] + K - 1 < lenS - (rpos[res] + K - 1) + 1) {
      printf("Yes\n%d %d\n", lpos[i], lenS - (rpos[res] + K - 1) + 1);
      return 0;
    }
  }
  puts("No");
  return 0;
}
