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 N = 1e6 + 500;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
const long double eps = 0.00001;
const int P = 1336;
const int inf = 1e9;
int mi[N], ma[N];
vector<int> z_f(string s) {
  int n = s.size(), l = 0, r = 0;
  vector<int> z;
  z.resize(n);
  int uk = 0;
  for (int i = 1; i < n; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (s[z[i]] == s[i + z[i]]) {
      z[i]++;
    }
    if (i + z[i] - 1 > r) {
      r = i + z[i] - 1;
      l = i;
    }
  }
  return z;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  int n, m, k;
  cin >> n >> m >> k;
  string a, b;
  cin >> a >> b;
  vector<int> z1, z2;
  z1 = z_f(b + "#" + a);
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  z2 = z_f(b + "#" + a);
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  if (m <= k) {
    for (int i = b.size() + 1; i < a.size() + b.size() + 1; i++) {
      if (z1[i] == m) {
        int p = i - b.size();
        if (p > k && p + k <= n) {
          cout << "Yes" << '\n';
          cout << 1 << " " << p;
          return 0;
        }
        if (p + k > n) {
          int v2 = n - k + 1;
          if (v2 > k) cout << "Yes" << '\n' << 1 << " " << v2;
          return 0;
        }
        if (p + 2 * k - 1 <= n) {
          cout << "Yes" << '\n';
          cout << p << " " << p + k;
          return 0;
        }
      }
    }
  }
  for (int i = 0; i <= n + 1; i++) {
    mi[i] = inf;
    ma[i] = -1;
  }
  vector<pair<int, int> > v;
  set<int> st;
  int add = 0;
  int uk = 0;
  for (int i = b.size() + 1; i < a.size() + b.size() + 1; i++) {
    if (z1[i] == 0) continue;
    int c = i + z1[i] - 1 - b.size();
    if (c >= k && mi[z1[i]] == inf) v.push_back({z1[i], c - k + 1});
  }
  sort(v.rbegin(), v.rend());
  st.insert(inf);
  for (int i = n; i > 0; i--) {
    while (uk < v.size() && v[uk].first == i) {
      int c = v[uk].second + add;
      st.insert(c);
      uk++;
    }
    mi[i] = *st.upper_bound(add);
    mi[i] -= add;
    add++;
  }
  v.clear();
  st.clear();
  for (int i = b.size() + 1; i < a.size() + b.size() + 1; i++) {
    if (z2[i] == 0) continue;
    int c = i + z2[i] - 1;
    c = a.size() + b.size() - c + 1;
    if (c + k - 1 <= a.size()) v.push_back({z2[i], c});
  }
  sort(v.rbegin(), v.rend());
  uk = 0;
  add = 0;
  st.insert(inf);
  for (int i = n; i > 0; i--) {
    while (uk < v.size() && v[uk].first == i) {
      int c = v[uk].second - add;
      st.insert(-c);
      uk++;
    }
    int p = n - k + 1;
    ma[i] = -(*st.lower_bound(-p));
    ma[i] += add;
    add++;
  }
  for (int i = 1; i <= m; i++) {
    int c = m - i;
    if (i > k || c > k) continue;
    if (mi[i] + k - 1 < ma[c]) {
      cout << "Yes" << '\n';
      cout << mi[i] << " " << ma[c] << '\n';
      return 0;
    }
  }
  cout << "No";
}
