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 INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 1e6 + 7;
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = (long long)a * a % M)
    if (b & 1) ret = (long long)ret * a % M;
  return ret;
}
int n, m, k;
char A[maxn], B[maxn];
int fail[maxn];
int _L, _R;
void solve(int F[]) {
  int i, j;
  fail[1] = 0;
  for (i = 2; i <= m; i++) {
    j = fail[i - 1];
    while (j && B[j + 1] != B[i]) j = fail[j];
    if (B[j + 1] == B[i])
      fail[i] = j + 1;
    else
      fail[i] = 0;
  }
  j = 0;
  for (i = 1; i <= n; i++) {
    while (j && B[j + 1] != A[i]) j = fail[j];
    if (B[j + 1] == A[i]) j++;
    if (!F[j] && i - k + 1 > 0) F[j] = i;
    if (j == m) {
      _R = i;
      if (!_L) _L = i;
    }
  }
  for (i = m; i >= 1; i--)
    if (F[i] && fail[i]) {
      if (F[fail[i]] > F[i] || !F[fail[i]]) F[fail[i]] = F[i];
    }
}
int pre[maxn], suf[maxn];
int main() {
  int i;
  scanf("%d%d%d", &n, &m, &k);
  scanf("%s%s", A + 1, B + 1);
  solve(pre);
  if (m <= k && _L) {
    int L, R;
    L = max(1, _L - k + 1);
    R = n - k + 1;
    if (L + k <= R) {
      puts("Yes");
      printf("%d %d\n", L, R);
      return 0;
    }
    L = 1;
    R = min(n - k + 1, _R - m + 1);
    if (L + k <= R) {
      puts("Yes");
      printf("%d %d\n", L, R);
      return 0;
    }
  }
  reverse(A + 1, A + 1 + n);
  reverse(B + 1, B + 1 + m);
  solve(suf);
  for (i = max(1, m - k); i <= min(m, k); i++)
    if (pre[i] && suf[m - i]) {
      int L = pre[i] - k + 1, R = n - suf[m - i] + 1;
      if (L + k <= R) {
        puts("Yes");
        printf("%d %d\n", L, R);
        return 0;
      }
    }
  puts("No");
  return 0;
}
