#include <bits/stdc++.h>
using namespace std;
const int m1 = 998244353;
const int m2 = 977787779;
const int m3 = 950747059;
struct hsh {
  int x, y, z;
  hsh(int a = 0) : x(a % m1), y(a % m2), z(a % m3) {}
  hsh(int a, int b, int c) : x(a), y(b), z(c) {}
  hsh operator+(hsh b) {
    return hsh((x + b.x) % m1, (y + b.y) % m2, (z + b.z) % m3);
  }
  hsh operator-(hsh b) {
    return hsh((x - b.x + m1) % m1, (y - b.y + m2) % m2, (z - b.z + m3) % m3);
  }
  hsh operator*(hsh b) {
    return hsh((long long)x * b.x % m1, (long long)y * b.y % m2,
               (long long)z * b.z % m3);
  }
  bool operator<(hsh b) const {
    return make_tuple(x, y, z) < make_tuple(b.x, b.y, b.z);
  }
  bool operator==(hsh b) const {
    return make_tuple(x, y, z) == make_tuple(b.x, b.y, b.z);
  }
};
const int N = 200050;
const int L = 18;
struct EERTREE {
  int IMAG, ZERO, tsz, go[N][26], len[N], link[N][L], at, sz, nod[N], pos[N];
  char s[N];
  void init() {
    tsz = sz = 0;
    IMAG = ++tsz;
    link[IMAG][0] = IMAG;
    len[IMAG] = -1;
    ZERO = ++tsz;
    link[ZERO][0] = IMAG;
    len[ZERO] = 0;
    at = IMAG;
  }
  EERTREE() { init(); }
  void Grow(char c) {
    s[++sz] = c;
    while (c != s[sz - len[at] - 1]) at = link[at][0];
    if (!go[at][c - 'a']) {
      go[at][c - 'a'] = ++tsz;
      pos[tsz] = sz;
      len[tsz] = len[at] + 2;
      int suf = link[at][0];
      while (c != s[sz - len[suf] - 1]) suf = link[suf][0];
      if (len[tsz] == 1)
        link[tsz][0] = ZERO;
      else
        link[tsz][0] = go[suf][c - 'a'];
    }
    at = go[at][c - 'a'];
    nod[sz] = at;
  }
  void Build(char s[], int n) {
    init();
    for (int i = 1; i <= n; i++) Grow(s[i]);
    for (int j = 1; j < L; j++)
      for (int i = 1; i <= tsz; i++) link[i][j] = link[link[i][j - 1]][j - 1];
  }
  int Get(int u, int x) {
    if (len[u] <= x) return len[u];
    for (int i = L - 1; ~i; i--)
      if (len[link[u][i]] > x) u = link[u][i];
    return len[link[u][0]];
  }
  bool Check(int u, int x) {
    for (int i = L - 1; ~i; i--)
      if (len[link[u][i]] >= x) u = link[u][i];
    return len[u] == x;
  }
} FW[2], BW[2];
int Decompose(int t, int l, int r, int n) {
  int pre = BW[t].Get(BW[t].nod[n - l + 1], r - l + 1);
  if (pre > 0 && FW[t].Check(FW[t].nod[r], r - l + 1 - pre)) return pre;
  int suf = FW[t].Get(FW[t].nod[r], r - l + 1);
  if (suf > 0 && BW[t].Check(BW[t].nod[n - l + 1], r - l + 1 - suf))
    return r - l + 1 - suf;
  return -1;
}
char s[N], t[N];
hsh H[2][N], po[N];
hsh Get(int t, int l, int r) { return H[t][r] - H[t][l - 1] * po[r - l + 1]; }
map<hsh, int> cnt[2];
set<hsh> st[2];
int main() {
  int n, m;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  FW[0].Build(s, n);
  FW[1].Build(t, m);
  reverse(s + 1, s + 1 + n);
  reverse(t + 1, t + 1 + m);
  BW[0].Build(s, n);
  BW[1].Build(t, m);
  reverse(s + 1, s + 1 + n);
  reverse(t + 1, t + 1 + m);
  long long ans = (long long)(FW[0].tsz - 2) * (FW[1].tsz - 2);
  po[0] = hsh(1);
  po[1] = hsh(777);
  for (int i = 2; i < N; i++) po[i] = po[i - 1] * po[1];
  for (int i = 1; i <= n; i++)
    H[0][i] = H[0][i - 1] * po[1] + hsh(s[i] - 'a' + 1);
  for (int i = 1; i <= m; i++)
    H[1][i] = H[1][i - 1] * po[1] + hsh(t[i] - 'a' + 1);
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      st[t].insert(Get(t, FW[t].pos[i] - FW[t].len[i] + 1, FW[t].pos[i]));
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      if (FW[t].len[border] * 2 >= FW[t].len[i]) cnt[t][Get(t, l, r)]++;
    }
  }
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      ans -= cnt[t ^ 1][Get(t, l, r)];
      int pre = Decompose(t, l, r, t == 0 ? n : m);
      if (~pre && pre != r - l + 1) {
        if (t == 0) {
          if (st[t ^ 1].count(Get(t, l, l + pre - 1)) &&
              st[t ^ 1].count(Get(t, l, r) * po[pre] + Get(t, l, l + pre - 1)))
            ans--;
        } else {
          int suf = r - l + 1 - pre;
          if (st[t ^ 1].count(Get(t, r - suf + 1, r)) &&
              st[t ^ 1].count(Get(t, r - suf + 1, r) * po[r - l + 1] +
                              Get(t, l, r)))
            ans--;
        }
      }
    }
  }
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      if (FW[t].len[border] * 2 < FW[t].len[i]) cnt[t][Get(t, l, r)]++;
    }
  }
  for (auto p : cnt[0]) ans += (long long)p.second * cnt[1][p.first];
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

After learning some fancy algorithms about palindromes, Chouti found palindromes very interesting, so he wants to challenge you with this problem.

Chouti has got two strings A and B. Since he likes [palindromes](https://en.wikipedia.org/wiki/Palindrome), he would like to pick a as some non-empty palindromic substring of A and b as some non-empty palindromic substring of B. Concatenating them, he will get string ab.

Chouti thinks strings he could get this way are interesting, so he wants to know how many different strings he can get.

Input

The first line contains a single string A (1 ≤ |A| ≤ 2 ⋅ 10^5).

The second line contains a single string B (1 ≤ |B| ≤ 2 ⋅ 10^5).

Strings A and B contain only lowercase English letters.

Output

The first and only line should contain a single integer — the number of possible strings.

Examples

Input

aa
aba


Output

6


Input

aaba
abaa


Output

15

Note

In the first example, attainable strings are 

  * "a" + "a" = "aa", 
  * "aa" + "a" = "aaa", 
  * "aa" + "aba" = "aaaba", 
  * "aa" + "b" = "aab", 
  * "a" + "aba" = "aaba", 
  * "a" + "b" = "ab". 



In the second example, attainable strings are "aa", "aaa", "aaaa", "aaaba", "aab", "aaba", "ab", "abaa", "abaaa", "abaaba", "abab", "ba", "baa", "baba", "bb".

Notice that though "a"+"aa"="aa"+"a"="aaa", "aaa" will only be counted once.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int m1 = 998244353;
const int m2 = 977787779;
const int m3 = 950747059;
struct hsh {
  int x, y, z;
  hsh(int a = 0) : x(a % m1), y(a % m2), z(a % m3) {}
  hsh(int a, int b, int c) : x(a), y(b), z(c) {}
  hsh operator+(hsh b) {
    return hsh((x + b.x) % m1, (y + b.y) % m2, (z + b.z) % m3);
  }
  hsh operator-(hsh b) {
    return hsh((x - b.x + m1) % m1, (y - b.y + m2) % m2, (z - b.z + m3) % m3);
  }
  hsh operator*(hsh b) {
    return hsh((long long)x * b.x % m1, (long long)y * b.y % m2,
               (long long)z * b.z % m3);
  }
  bool operator<(hsh b) const {
    return make_tuple(x, y, z) < make_tuple(b.x, b.y, b.z);
  }
  bool operator==(hsh b) const {
    return make_tuple(x, y, z) == make_tuple(b.x, b.y, b.z);
  }
};
const int N = 200050;
const int L = 18;
struct EERTREE {
  int IMAG, ZERO, tsz, go[N][26], len[N], link[N][L], at, sz, nod[N], pos[N];
  char s[N];
  void init() {
    tsz = sz = 0;
    IMAG = ++tsz;
    link[IMAG][0] = IMAG;
    len[IMAG] = -1;
    ZERO = ++tsz;
    link[ZERO][0] = IMAG;
    len[ZERO] = 0;
    at = IMAG;
  }
  EERTREE() { init(); }
  void Grow(char c) {
    s[++sz] = c;
    while (c != s[sz - len[at] - 1]) at = link[at][0];
    if (!go[at][c - 'a']) {
      go[at][c - 'a'] = ++tsz;
      pos[tsz] = sz;
      len[tsz] = len[at] + 2;
      int suf = link[at][0];
      while (c != s[sz - len[suf] - 1]) suf = link[suf][0];
      if (len[tsz] == 1)
        link[tsz][0] = ZERO;
      else
        link[tsz][0] = go[suf][c - 'a'];
    }
    at = go[at][c - 'a'];
    nod[sz] = at;
  }
  void Build(char s[], int n) {
    init();
    for (int i = 1; i <= n; i++) Grow(s[i]);
    for (int j = 1; j < L; j++)
      for (int i = 1; i <= tsz; i++) link[i][j] = link[link[i][j - 1]][j - 1];
  }
  int Get(int u, int x) {
    if (len[u] <= x) return len[u];
    for (int i = L - 1; ~i; i--)
      if (len[link[u][i]] > x) u = link[u][i];
    return len[link[u][0]];
  }
  bool Check(int u, int x) {
    for (int i = L - 1; ~i; i--)
      if (len[link[u][i]] >= x) u = link[u][i];
    return len[u] == x;
  }
} FW[2], BW[2];
int Decompose(int t, int l, int r, int n) {
  int pre = BW[t].Get(BW[t].nod[n - l + 1], r - l + 1);
  if (pre > 0 && FW[t].Check(FW[t].nod[r], r - l + 1 - pre)) return pre;
  int suf = FW[t].Get(FW[t].nod[r], r - l + 1);
  if (suf > 0 && BW[t].Check(BW[t].nod[n - l + 1], r - l + 1 - suf))
    return r - l + 1 - suf;
  return -1;
}
char s[N], t[N];
hsh H[2][N], po[N];
hsh Get(int t, int l, int r) { return H[t][r] - H[t][l - 1] * po[r - l + 1]; }
map<hsh, int> cnt[2];
set<hsh> st[2];
int main() {
  int n, m;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  FW[0].Build(s, n);
  FW[1].Build(t, m);
  reverse(s + 1, s + 1 + n);
  reverse(t + 1, t + 1 + m);
  BW[0].Build(s, n);
  BW[1].Build(t, m);
  reverse(s + 1, s + 1 + n);
  reverse(t + 1, t + 1 + m);
  long long ans = (long long)(FW[0].tsz - 2) * (FW[1].tsz - 2);
  po[0] = hsh(1);
  po[1] = hsh(777);
  for (int i = 2; i < N; i++) po[i] = po[i - 1] * po[1];
  for (int i = 1; i <= n; i++)
    H[0][i] = H[0][i - 1] * po[1] + hsh(s[i] - 'a' + 1);
  for (int i = 1; i <= m; i++)
    H[1][i] = H[1][i - 1] * po[1] + hsh(t[i] - 'a' + 1);
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      st[t].insert(Get(t, FW[t].pos[i] - FW[t].len[i] + 1, FW[t].pos[i]));
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      if (FW[t].len[border] * 2 >= FW[t].len[i]) cnt[t][Get(t, l, r)]++;
    }
  }
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      ans -= cnt[t ^ 1][Get(t, l, r)];
      int pre = Decompose(t, l, r, t == 0 ? n : m);
      if (~pre && pre != r - l + 1) {
        if (t == 0) {
          if (st[t ^ 1].count(Get(t, l, l + pre - 1)) &&
              st[t ^ 1].count(Get(t, l, r) * po[pre] + Get(t, l, l + pre - 1)))
            ans--;
        } else {
          int suf = r - l + 1 - pre;
          if (st[t ^ 1].count(Get(t, r - suf + 1, r)) &&
              st[t ^ 1].count(Get(t, r - suf + 1, r) * po[r - l + 1] +
                              Get(t, l, r)))
            ans--;
        }
      }
    }
  }
  for (int t = 0; t < 2; t++) {
    for (int i = 3; i <= FW[t].tsz; i++) {
      int border = FW[t].link[i][0], l, r;
      if (border < 3) continue;
      if (t == 0)
        l = FW[t].pos[i] - (FW[t].len[i] - FW[t].len[border]) + 1,
        r = FW[t].pos[i];
      else
        l = FW[t].pos[i] - FW[t].len[i] + 1,
        r = FW[t].pos[i] - FW[t].len[border];
      if (FW[t].len[border] * 2 < FW[t].len[i]) cnt[t][Get(t, l, r)]++;
    }
  }
  for (auto p : cnt[0]) ans += (long long)p.second * cnt[1][p.first];
  printf("%lld\n", ans);
  return 0;
}

```