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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 998244353;
const int N = 500005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
int nxt[2][N], f[N], g[N], n, m, k;
char s[N], t[N];
bool v[N], ok = false;
void kmp_pre(char st[], int m, int nxt[]) {
  int i = 0, j = nxt[0] = -1;
  while (i < m) {
    while (j != -1 && st[i] != st[j]) j = nxt[j];
    nxt[++i] = ++j;
  }
}
void work(int flag, int f[]) {
  if (flag) reverse(s, s + n), reverse(t, t + m);
  kmp_pre(t, m, nxt[flag]), memset(v, 0, sizeof(v)), v[0] = 1;
  int L = 0;
  for (int i = (0); i <= (n - 1); ++i) {
    while (L != -1 && s[i] != t[L]) L = nxt[flag][L];
    L++;
    if (L >= m) ok = true;
    if (i >= k - 1)
      for (int t = L; !v[t]; t = nxt[flag][t]) f[t - 1] = i, v[t] = 1;
  }
  if (flag) reverse(f, f + m);
  if (flag)
    for (int i = (0); i <= (m - 1); ++i)
      if (f[i] != -1) f[i] = n - 1 - f[i];
}
int main() {
  scanf("%d%d%d%s%s", &n, &m, &k, s, t), memset(f, -1, sizeof(f)),
      memset(g, -1, sizeof(g)), work(0, f), work(1, g);
  for (int i = (0); i <= (m - 2); ++i)
    if (i + 1 <= k && m - 1 - i <= k && f[i] != -1 && g[i + 1] != -1 &&
        f[i] < g[i + 1])
      return 0 * printf("Yes\n%d %d\n", f[i] - k + 2, g[i + 1] + 1);
  if (m <= k && g[0] >= k) return 0 * printf("Yes\n%d %d\n", 1, g[0] + 1);
  if (m <= k && f[m - 1] != -1 && f[m - 1] <= n - k - 1)
    return 0 * printf("Yes\n%d %d\n", f[m - 1] - k + 2, n - k);
  if (ok) return 0 * printf("Yes\n%d %d\n", 1, n - k + 1);
  printf("No\n");
  return 0;
}
