#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int mod1 = 998244353;
const int mod2 = 998244853;
const int base = 131;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    o = -1, ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0', ch = getchar();
  }
  return x * o;
}
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return make_pair((a.first + b.first) % mod1, (a.second + b.second) % mod2);
}
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
  return make_pair((a.first - b.first + mod1) % mod1,
                   (a.second - b.second + mod2) % mod2);
}
pair<int, int> operator*(pair<int, int> a, pair<int, int> b) {
  return make_pair(1ll * a.first * b.first % mod1,
                   1ll * a.second * b.second % mod2);
}
pair<int, int> operator*(pair<int, int> a, int b) {
  return make_pair(1ll * a.first * b % mod1, 1ll * a.second * b % mod2);
}
pair<int, int> operator+(pair<int, int> a, int b) {
  return make_pair((a.first + b) % mod1, (a.second + b) % mod2);
}
int n, m;
char a[N], b[N];
pair<int, int> ha[N], hb[N], pw[N];
long long ans = 0;
struct PAM {
  int ch[N][26], fa[N], len[N], tot = 1, last = 1, anc[20][N], pos[N], r[N];
  char s[N];
  void extend(int n, int c) {
    int p = last;
    while (s[n - len[p] - 1] != s[n]) {
      p = fa[p];
    }
    if (!ch[p][c]) {
      int v = ++tot;
      len[v] = len[p] + 2;
      int k = fa[p];
      while (s[n - len[k] - 1] != s[n]) {
        k = fa[k];
      }
      fa[v] = ch[k][c];
      ch[p][c] = v;
    }
    last = ch[p][c];
    pos[n] = last;
    r[last] = n;
  }
  void build(char *t) {
    int n = strlen(t + 1);
    for (int i = 0; i <= n + 1; i++) {
      s[i] = t[i];
    }
    fa[0] = fa[1] = 1, len[1] = -1;
    for (int i = 1; i <= n; i++) {
      extend(i, s[i] - 'a');
    }
    for (int i = 0; i <= tot; i++) {
      anc[0][i] = fa[i];
    }
    for (int i = 1; i <= 18; i++)
      for (int j = 0; j <= tot; j++) {
        anc[i][j] = anc[i - 1][anc[i - 1][j]];
      }
  }
  int jump(int r, int l) {
    int u = pos[r];
    if (len[u] <= l) {
      return len[u];
    }
    for (int i = 18; ~i; i--)
      if (len[anc[i][u]] > l) {
        u = anc[i][u];
      }
    return len[fa[u]];
  }
  bool check(int r, int l) { return jump(r, l) == l; }
} A, B, RA, RB;
map<pair<int, int>, int> fa, fb, ga, gb;
pair<int, int> gethsh(pair<int, int> *hsh, int l, int r) {
  return hsh[r] - hsh[l - 1] * pw[r - l + 1];
}
int main() {
  scanf("%s", a + 1), scanf("%s", b + 1);
  n = strlen(a + 1), m = strlen(b + 1);
  pw[0] = make_pair(1, 1);
  for (int i = 1; i <= n || i <= m; i++) {
    pw[i] = pw[i - 1] * base;
  }
  for (int i = 1; i <= n; i++) {
    ha[i] = ha[i - 1] * base + (a[i] - 'a' + 1);
  }
  for (int i = 1; i <= m; i++) {
    hb[i] = hb[i - 1] * base + (b[i] - 'a' + 1);
  }
  A.build(a), B.build(b);
  reverse(a + 1, a + n + 1), reverse(b + 1, b + m + 1);
  RA.build(a), RB.build(b);
  ans = 1ll * (A.tot - 1) * (B.tot - 1);
  for (int i = 2; i <= A.tot; i++) {
    ++fa[gethsh(ha, A.r[i] - A.len[i] + 1, A.r[i])];
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    if (A.len[i] <= (A.len[p] << 1)) {
      ++ga[gethsh(ha, A.r[i] - (A.len[i] - A.len[p]) + 1, A.r[i])];
    }
  }
  for (int i = 2; i <= B.tot; i++) {
    ++fb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i])];
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    if (B.len[i] <= (B.len[p] << 1)) {
      ++gb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i] - B.len[p])];
    }
  }
  for (int i = 2; i <= A.tot; i++) {
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    int l = A.r[i] - (A.len[i] - A.len[p]) + 1, r = A.r[i];
    ans -= gb[gethsh(ha, l, r)];
    int lps = A.jump(r, r - l + 1);
    if (lps == r - l + 1) {
      continue;
    }
    if (RA.check(n - l + 1, r - l + 1 - lps)) {
      if (fb.count(gethsh(ha, l, r) * pw[r - l + 1 - lps] +
                   gethsh(ha, l, r - lps))) {
        --ans;
      }
      continue;
    }
    int lpp = RA.jump(n - l + 1, r - l + 1);
    if (A.check(r, r - l + 1 - lpp)) {
      if (fb.count(gethsh(ha, l, r) * pw[lpp] + gethsh(ha, l, l + lpp - 1))) {
        --ans;
      }
    }
  }
  for (int i = 2; i <= B.tot; i++) {
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    int l = B.r[i] - B.len[i] + 1, r = B.r[i] - B.len[p];
    ans -= ga[gethsh(hb, l, r)];
    int lps = B.jump(r, r - l + 1);
    if (lps == r - l + 1) {
      continue;
    }
    if (RB.check(m - l + 1, r - l + 1 - lps)) {
      if (fa.count(gethsh(hb, r - lps + 1, r) * pw[r - l + 1] +
                   gethsh(hb, l, r))) {
        --ans;
      }
      continue;
    }
    int lpp = RB.jump(m - l + 1, r - l + 1);
    if (B.check(r, r - l + 1 - lpp)) {
      if (fa.count(gethsh(hb, l + lpp, r) * pw[r - l + 1] + gethsh(hb, l, r))) {
        --ans;
      }
    }
  }
  ga.clear(), gb.clear();
  for (int i = 2; i <= A.tot; i++) {
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    ++ga[gethsh(ha, A.r[i] - (A.len[i] - A.len[p]) + 1, A.r[i])];
  }
  for (int i = 2; i <= B.tot; i++) {
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    ++gb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i] - B.len[p])];
  }
  for (auto x : ga) {
    ans += 1ll * x.second * gb[x.first];
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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 N = 2e5 + 10;
const int mod1 = 998244353;
const int mod2 = 998244853;
const int base = 131;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    o = -1, ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0', ch = getchar();
  }
  return x * o;
}
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return make_pair((a.first + b.first) % mod1, (a.second + b.second) % mod2);
}
pair<int, int> operator-(pair<int, int> a, pair<int, int> b) {
  return make_pair((a.first - b.first + mod1) % mod1,
                   (a.second - b.second + mod2) % mod2);
}
pair<int, int> operator*(pair<int, int> a, pair<int, int> b) {
  return make_pair(1ll * a.first * b.first % mod1,
                   1ll * a.second * b.second % mod2);
}
pair<int, int> operator*(pair<int, int> a, int b) {
  return make_pair(1ll * a.first * b % mod1, 1ll * a.second * b % mod2);
}
pair<int, int> operator+(pair<int, int> a, int b) {
  return make_pair((a.first + b) % mod1, (a.second + b) % mod2);
}
int n, m;
char a[N], b[N];
pair<int, int> ha[N], hb[N], pw[N];
long long ans = 0;
struct PAM {
  int ch[N][26], fa[N], len[N], tot = 1, last = 1, anc[20][N], pos[N], r[N];
  char s[N];
  void extend(int n, int c) {
    int p = last;
    while (s[n - len[p] - 1] != s[n]) {
      p = fa[p];
    }
    if (!ch[p][c]) {
      int v = ++tot;
      len[v] = len[p] + 2;
      int k = fa[p];
      while (s[n - len[k] - 1] != s[n]) {
        k = fa[k];
      }
      fa[v] = ch[k][c];
      ch[p][c] = v;
    }
    last = ch[p][c];
    pos[n] = last;
    r[last] = n;
  }
  void build(char *t) {
    int n = strlen(t + 1);
    for (int i = 0; i <= n + 1; i++) {
      s[i] = t[i];
    }
    fa[0] = fa[1] = 1, len[1] = -1;
    for (int i = 1; i <= n; i++) {
      extend(i, s[i] - 'a');
    }
    for (int i = 0; i <= tot; i++) {
      anc[0][i] = fa[i];
    }
    for (int i = 1; i <= 18; i++)
      for (int j = 0; j <= tot; j++) {
        anc[i][j] = anc[i - 1][anc[i - 1][j]];
      }
  }
  int jump(int r, int l) {
    int u = pos[r];
    if (len[u] <= l) {
      return len[u];
    }
    for (int i = 18; ~i; i--)
      if (len[anc[i][u]] > l) {
        u = anc[i][u];
      }
    return len[fa[u]];
  }
  bool check(int r, int l) { return jump(r, l) == l; }
} A, B, RA, RB;
map<pair<int, int>, int> fa, fb, ga, gb;
pair<int, int> gethsh(pair<int, int> *hsh, int l, int r) {
  return hsh[r] - hsh[l - 1] * pw[r - l + 1];
}
int main() {
  scanf("%s", a + 1), scanf("%s", b + 1);
  n = strlen(a + 1), m = strlen(b + 1);
  pw[0] = make_pair(1, 1);
  for (int i = 1; i <= n || i <= m; i++) {
    pw[i] = pw[i - 1] * base;
  }
  for (int i = 1; i <= n; i++) {
    ha[i] = ha[i - 1] * base + (a[i] - 'a' + 1);
  }
  for (int i = 1; i <= m; i++) {
    hb[i] = hb[i - 1] * base + (b[i] - 'a' + 1);
  }
  A.build(a), B.build(b);
  reverse(a + 1, a + n + 1), reverse(b + 1, b + m + 1);
  RA.build(a), RB.build(b);
  ans = 1ll * (A.tot - 1) * (B.tot - 1);
  for (int i = 2; i <= A.tot; i++) {
    ++fa[gethsh(ha, A.r[i] - A.len[i] + 1, A.r[i])];
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    if (A.len[i] <= (A.len[p] << 1)) {
      ++ga[gethsh(ha, A.r[i] - (A.len[i] - A.len[p]) + 1, A.r[i])];
    }
  }
  for (int i = 2; i <= B.tot; i++) {
    ++fb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i])];
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    if (B.len[i] <= (B.len[p] << 1)) {
      ++gb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i] - B.len[p])];
    }
  }
  for (int i = 2; i <= A.tot; i++) {
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    int l = A.r[i] - (A.len[i] - A.len[p]) + 1, r = A.r[i];
    ans -= gb[gethsh(ha, l, r)];
    int lps = A.jump(r, r - l + 1);
    if (lps == r - l + 1) {
      continue;
    }
    if (RA.check(n - l + 1, r - l + 1 - lps)) {
      if (fb.count(gethsh(ha, l, r) * pw[r - l + 1 - lps] +
                   gethsh(ha, l, r - lps))) {
        --ans;
      }
      continue;
    }
    int lpp = RA.jump(n - l + 1, r - l + 1);
    if (A.check(r, r - l + 1 - lpp)) {
      if (fb.count(gethsh(ha, l, r) * pw[lpp] + gethsh(ha, l, l + lpp - 1))) {
        --ans;
      }
    }
  }
  for (int i = 2; i <= B.tot; i++) {
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    int l = B.r[i] - B.len[i] + 1, r = B.r[i] - B.len[p];
    ans -= ga[gethsh(hb, l, r)];
    int lps = B.jump(r, r - l + 1);
    if (lps == r - l + 1) {
      continue;
    }
    if (RB.check(m - l + 1, r - l + 1 - lps)) {
      if (fa.count(gethsh(hb, r - lps + 1, r) * pw[r - l + 1] +
                   gethsh(hb, l, r))) {
        --ans;
      }
      continue;
    }
    int lpp = RB.jump(m - l + 1, r - l + 1);
    if (B.check(r, r - l + 1 - lpp)) {
      if (fa.count(gethsh(hb, l + lpp, r) * pw[r - l + 1] + gethsh(hb, l, r))) {
        --ans;
      }
    }
  }
  ga.clear(), gb.clear();
  for (int i = 2; i <= A.tot; i++) {
    int p = A.fa[i];
    if (p < 2) {
      continue;
    }
    ++ga[gethsh(ha, A.r[i] - (A.len[i] - A.len[p]) + 1, A.r[i])];
  }
  for (int i = 2; i <= B.tot; i++) {
    int p = B.fa[i];
    if (p < 2) {
      continue;
    }
    ++gb[gethsh(hb, B.r[i] - B.len[i] + 1, B.r[i] - B.len[p])];
  }
  for (auto x : ga) {
    ans += 1ll * x.second * gb[x.first];
  }
  cout << ans;
  return 0;
}

```