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 long long INF = 1e9;
inline long long read() {
  long long X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return w ? -X : X;
}
long long hsh = 29;
long long mod = 1000000007;
long long n, m, k;
char s[500005], t[5000005];
long long na[500005], pa[500005];
long long hashs[500005];
long long hasht[500005];
long long p[500005];
bool check(long long l, long long r, long long L, long long R) {
  return (((hashs[r] - 1LL * hashs[l - 1] * p[r - l + 1]) % mod + mod) % mod) ==
         ((mod + (hasht[R] - 1LL * hasht[L - 1] * p[R - L + 1]) % mod) % mod);
}
void solve1(long long l, long long r) {
  long long pos = k;
  for (long long i = l; i <= r; i++) {
    while (pos < i) pos++;
    while (pos <= n) {
      if (check(pos - i + 1, pos, 1, i)) break;
      pos++;
    }
    if (pos <= n) {
      na[i] = pos;
    } else {
      na[i] = 0;
    }
  }
}
void solve2(long long l, long long r) {
  long long pos = n - k + 1;
  for (long long i = r; i >= l; i--) {
    while (pos + m - i > n) {
      pos--;
    }
    while (pos) {
      if (check(pos, pos + m - i, i, m)) break;
      pos--;
    }
    pa[i] = pos;
  }
}
void special() {
  long long pos = 1;
  for (long long i = 1; i <= m; i++) {
    while (pos < i) pos++;
    while (pos <= n) {
      if (check(pos - i + 1, pos, 1, i)) {
        break;
      }
      pos++;
    }
  }
  if (max(pos, k) + k <= n) {
    cout << "Yes" << endl;
    cout << max(1LL, pos - k + 1) << " " << max(pos, k) + 1 << endl;
    exit(0);
  }
  pos = n;
  for (long long i = m; i; i--) {
    while (pos + m - i > n) pos--;
    while (pos) {
      if (check(pos, pos + m - i, i, m)) {
        break;
      }
      pos--;
    }
  }
  if (n - max(n - pos + 1, k) - k > 0) {
    cout << "Yes" << endl;
    cout << n - max(n - pos + 1, k) - k << " " << n - max(n - pos + 1, k) + 1;
    exit(0);
  }
}
int main() {
  cin >> n >> m >> k;
  scanf("%s%s", s + 1, t + 1);
  p[0] = 1;
  for (long long i = 1; i <= n; i++) {
    p[i] = (p[i - 1] * hsh) % mod;
  }
  for (long long i = 1; i <= n; i++) {
    hashs[i] = (hashs[i - 1] * hsh + s[i] - 'a') % mod;
  }
  for (long long i = 1; i <= m; i++) {
    hasht[i] = (hasht[i - 1] * hsh + t[i] - 'a') % mod;
  }
  if (m <= k) {
    special();
  }
  long long tmp1 = 1;
  long long tmp2 = 1;
  for (long long i = 1; i <= m and i <= k; i++) {
    if (check(k - i + 1, k, 1, i)) {
      tmp1 = i;
      if (tmp2 != 1) {
        tmp2 = i;
      }
      na[i] = k;
    }
  }
  solve1(1, tmp2 - 1);
  solve1(tmp1 + 1, min(m, k));
  tmp1 = tmp2 = m;
  for (long long i = m; i and i >= m - k + 1; i--) {
    if (check(n - k + 1, n - k + 1 + m - i, i, m)) {
      tmp1 = i;
      if (tmp2 != m) {
        tmp2 = i;
      }
      pa[i] = n - k + 1;
    }
  }
  solve2(tmp2 + 1, m);
  solve2(max(1LL, m - k + 1), tmp1 - 1);
  long long x = 0, y = 0;
  for (long long i = 1; i < m; i++) {
    if (na[i] and pa[i + 1] and na[i] < pa[i + 1]) {
      x = na[i] - k + 1;
      y = pa[i + 1];
      break;
    }
  }
  if (x) {
    cout << "Yes" << endl;
    cout << x << " " << y << endl;
    return 0;
  }
  cout << "No";
  return 0;
}
