You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D. 

You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.

Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B. 

If X is a string, |X| denotes its length.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters. 

Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement. 

You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B. 

The second line contains a string consisting of n lowercase Latin letters — string A.

The third line contains a string consisting of m lowercase Latin letters — string B. 

Output

Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B. 

Examples

Input


4 5
abba
babab


Output


5

Input


8 10
bbbbabab
bbbabaaaaa


Output


12

Input


7 7
uiibwws
qhtkxcn


Output


0

Note

For the first case:

abb from the first string and abab from the second string have LCS equal to abb.

The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
#include <bits/stdc++.h>
using namespace std;
const long long INF(0x3f3f3f3f3f3f3f3fll);
const long long inf(0x3f3f3f3f);
template <typename T>
void read(T &res) {
  bool flag = false;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
  for (res = ch - 48; isdigit(ch = getchar());
       res = (res << 1) + (res << 3) + ch - 48)
    ;
  flag && (res = -res);
}
template <typename T>
void Out(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) Out(x / 10);
  putchar(x % 10 + '0');
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long pow_mod(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long fact_pow(long long n, long long p) {
  long long res = 0;
  while (n) {
    n /= p;
    res += n;
  }
  return res;
}
long long mult(long long a, long long b, long long p) {
  a %= p;
  b %= p;
  long long r = 0, v = a;
  while (b) {
    if (b & 1) {
      r += v;
      if (r > p) r -= p;
    }
    v <<= 1;
    if (v > p) v -= p;
    b >>= 1;
  }
  return r;
}
long long quick_pow(long long a, long long b, long long p) {
  long long r = 1, v = a % p;
  while (b) {
    if (b & 1) r = mult(r, v, p);
    v = mult(v, v, p);
    b >>= 1;
  }
  return r;
}
bool CH(long long a, long long n, long long x, long long t) {
  long long r = quick_pow(a, x, n);
  long long z = r;
  for (long long i = 1; i <= t; i++) {
    r = mult(r, r, n);
    if (r == 1 && z != 1 && z != n - 1) return true;
    z = r;
  }
  return r != 1;
}
bool Miller_Rabin(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (!(n & 1)) return false;
  long long x = n - 1, t = 0;
  while (!(x & 1)) {
    x >>= 1;
    t++;
  }
  srand(time(NULL));
  long long o = 8;
  for (long long i = 0; i < o; i++) {
    long long a = rand() % (n - 1) + 1;
    if (CH(a, n, x, t)) return false;
  }
  return true;
}
long long exgcd1(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long t = exgcd1(b, a % b, y, x);
  y -= a / b * x;
  return t;
}
long long get_inv(long long a, long long mod) {
  long long x, y;
  long long d = exgcd1(a, mod, x, y);
  return d == 1 ? (x % mod + mod) % mod : -1;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long t = x;
  x = y, y = t - (a / b) * y;
}
long long INV(long long a, long long b) {
  long long x, y;
  return exgcd(a, b, x, y), (x % b + b) % b;
}
long long crt(long long x, long long p, long long mod) {
  return INV(p / mod, mod) * (p / mod) * x;
}
long long FAC(long long x, long long a, long long b) {
  if (!x) return 1;
  long long ans = 1;
  for (long long i = 1; i <= b; i++)
    if (i % a) ans *= i, ans %= b;
  ans = pow_mod(ans, x / b, b);
  for (long long i = 1; i <= x % b; i++)
    if (i % a) ans *= i, ans %= b;
  return ans * FAC(x / a, a, b) % b;
}
long long C(long long n, long long m, long long a, long long b) {
  long long N = FAC(n, a, b), M = FAC(m, a, b), Z = FAC(n - m, a, b), sum = 0,
            i;
  for (i = n; i; i = i / a) sum += i / a;
  for (i = m; i; i = i / a) sum -= i / a;
  for (i = n - m; i; i = i / a) sum -= i / a;
  return N * pow_mod(a, sum, b) % b * INV(M, b) % b * INV(Z, b) % b;
}
long long exlucas(long long n, long long m, long long p) {
  long long t = p, ans = 0, i;
  for (i = 2; i * i <= p; i++) {
    long long k = 1;
    while (t % i == 0) {
      k *= i, t /= i;
    }
    ans += crt(C(n, m, i, k), p, k), ans %= p;
  }
  if (t > 1) ans += crt(C(n, m, t, t), p, t), ans %= p;
  return ans % p;
}
long long prime[100010], minprime[100010], phi[100010];
long long euler(long long n) {
  long long c = 0, i, j;
  phi[1] = 1;
  for (i = 2; i <= n; i++) {
    if (!minprime[i]) prime[++c] = i, minprime[i] = i, phi[i] = i - 1;
    for (j = 1; j <= c && i * prime[j] <= n; j++) {
      minprime[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      } else
        phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  return c;
}
struct Tree {
  long long l, r, sum, lazy, maxn, minn;
} tree[100000];
inline void push_up(long long rt) {
  tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
  tree[rt].maxn = max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn);
  tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
inline void push_down(long long rt, long long length) {
  if (tree[rt].lazy) {
    tree[rt << 1].lazy += tree[rt].lazy;
    tree[rt << 1 | 1].lazy += tree[rt].lazy;
    tree[rt << 1].sum += (length - (length >> 1)) * tree[rt].lazy;
    tree[rt << 1 | 1].sum += (length >> 1) * tree[rt].lazy;
    tree[rt << 1].minn += tree[rt].lazy;
    tree[rt << 1 | 1].minn += tree[rt].lazy;
    tree[rt << 1].maxn += tree[rt].lazy;
    tree[rt << 1 | 1].maxn += tree[rt].lazy;
    tree[rt].lazy = 0;
  }
}
inline void build(long long l, long long r, long long rt, long long *aa) {
  tree[rt].lazy = 0;
  tree[rt].l = l;
  tree[rt].r = r;
  if (l == r) {
    tree[rt].sum = aa[l];
    tree[rt].minn = tree[rt].sum;
    tree[rt].maxn = tree[rt].sum;
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, rt << 1, aa);
  build(mid + 1, r, rt << 1 | 1, aa);
  push_up(rt);
}
inline void update_range(long long L, long long R, long long key,
                         long long rt) {
  if (tree[rt].r < L || tree[rt].l > R) return;
  if (L <= tree[rt].l && R >= tree[rt].r) {
    tree[rt].sum += (tree[rt].r - tree[rt].l + 1) * key;
    tree[rt].minn += key;
    tree[rt].maxn += key;
    tree[rt].lazy += key;
    return;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  if (L <= mid) update_range(L, R, key, rt << 1);
  if (R > mid) update_range(L, R, key, rt << 1 | 1);
  push_up(rt);
}
inline long long query_range(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].sum;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = 0;
  if (L <= mid) ans += query_range(L, R, rt << 1);
  if (R > mid) ans += query_range(L, R, rt << 1 | 1);
  return ans;
}
inline long long query_min(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].minn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = (0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = min(ans, query_min(L, R, rt << 1));
  if (R > mid) ans = min(ans, query_min(L, R, rt << 1 | 1));
  return ans;
}
inline long long query_max(long long L, long long R, long long rt) {
  if (L <= tree[rt].l && R >= tree[rt].r) {
    return tree[rt].maxn;
  }
  push_down(rt, tree[rt].r - tree[rt].l + 1);
  long long mid = (tree[rt].r + tree[rt].l) >> 1;
  long long ans = -(0x3f3f3f3f3f3f3f3fll);
  if (L <= mid) ans = max(ans, query_max(L, R, rt << 1));
  if (R > mid) ans = max(ans, query_max(L, R, rt << 1 | 1));
  return ans;
}
namespace linear_seq {
const long long N = 1e5 + 10, mod = 1e9 + 7;
long long Rx[N], Ba[N], _c[N], _md[N];
vector<long long> Md;
long long PMD(long long a, long long b) {
  long long Rx = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) Rx = Rx * a % mod;
    a = a * a % mod;
  }
  return Rx;
}
void MUL(long long *a, long long *b, long long k) {
  for (long long i = 0; i < k + k; i++) _c[i] = 0;
  for (long long i = 0; i < k; i++)
    if (a[i])
      for (long long j = 0; j < k; j++)
        _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
  for (long long i = k + k - 1; i >= k; i--)
    if (_c[i])
      for (long long j = 0; j < ((long long)(Md).size()); j++)
        _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
  for (long long i = 0; i < k; i++) a[i] = _c[i];
}
long long Solve(long long n, vector<long long> a, vector<long long> b) {
  long long ans = 0, pnt = 0, k = a.size();
  assert(((long long)(a).size()) == ((long long)(b).size()));
  for (long long i = 0; i < k; i++) _md[k - 1 - i] = -a[i];
  _md[k] = 1;
  Md.clear();
  for (long long i = 0; i < k; i++)
    if (_md[i] != 0) Md.push_back(i);
  for (long long i = 0; i < k; i++) Rx[i] = Ba[i] = 0;
  Rx[0] = 1;
  while ((1ll << pnt) <= n) pnt++;
  for (long long p = pnt; p >= 0; p--) {
    MUL(Rx, Rx, k);
    if ((n >> p) & 1) {
      for (long long i = k - 1; i >= 0; i--) Rx[i + 1] = Rx[i];
      Rx[0] = 0;
      for (long long j = 0; j < ((long long)(Md).size()); j++)
        Rx[Md[j]] = (Rx[Md[j]] - Rx[k] * _md[Md[j]]) % mod;
    }
  }
  for (long long i = 0; i < k; i++) ans = (ans + Rx[i] * b[i]) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
vector<long long> BM(vector<long long> s) {
  vector<long long> C(1, 1), B(1, 1);
  long long L = 0, m = 1, b = 1;
  for (long long n = 0; n < ((long long)(s).size()); n++) {
    long long d = 0;
    for (long long i = 0; i < L + 1; i++)
      d = (d + (long long)C[i] * s[n - i]) % mod;
    if (d == 0)
      ++m;
    else if (2 * L <= n) {
      vector<long long> T = C;
      long long c = mod - d * PMD(b, mod - 2) % mod;
      while (((long long)(C).size()) < ((long long)(B).size()) + m)
        C.push_back(0);
      for (long long i = 0; i < ((long long)(B).size()); i++)
        C[i + m] = (C[i + m] + c * B[i]) % mod;
      L = n + 1 - L;
      B = T;
      b = d;
      m = 1;
    } else {
      long long c = mod - d * PMD(b, mod - 2) % mod;
      while (((long long)(C).size()) < ((long long)(B).size()) + m)
        C.push_back(0);
      for (long long i = 0; i < ((long long)(B).size()); i++)
        C[i + m] = (C[i + m] + c * B[i]) % mod;
      ++m;
    }
  }
  return C;
}
long long Gao(vector<long long> a, long long n) {
  vector<long long> c = BM(a);
  c.erase(c.begin());
  n--;
  for (long long i = 0; i < ((long long)(c).size()); i++)
    c[i] = (mod - c[i]) % mod;
  return Solve(
      n, c, vector<long long>(a.begin(), a.begin() + ((long long)(c).size())));
}
};  // namespace linear_seq
const long long N = 3e5 + 10;
struct Edge {
  long long nex, to;
} edge[N << 1];
long long head[N], TOT;
void add_edge(long long u, long long v) {
  edge[++TOT].nex = head[u];
  edge[TOT].to = v;
  head[u] = TOT;
}
const long long M = 5e3 + 10;
char s[N], t[N];
long long dp[M][M];
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n, m, ans = 0;
  cin >> n >> m >> s + 1 >> t + 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) {
      if (s[i] == t[j])
        dp[i][j] = max(max(2LL, dp[i][j]), dp[i - 1][j - 1] + 2);
      else
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] - 2);
      dp[i][j] = max(dp[i][j], dp[i][j - 1] - 1);
      dp[i][j] = max(dp[i][j], dp[i - 1][j] - 1);
      ans = max(ans, dp[i][j]);
    }
  cout << ans << '\n';
  return 0;
}
