//by 72
#include<bits/stdc++.h>
#define F(i, a, b) for(int i = a; i <= b; i ++)
#define Fd(i, a, b) for(int i = a; i >= b; i --)
#define pb push_back
#define pii pair<int, int>
#define fi first
#define se second
#define int long long
using namespace std;

const int mod = 998244353;
const int N = 5010;
const int inf = 1e18;
typedef array<int, 3> a3; 
typedef long long ll;

void exkmp(string a, string b, vector<int> &z, vector<int> &p) {
    int n = a.size(), m = b.size(), now = 0;
    a = " " + a, b = " " + b; 
    z.clear(), p.clear();
    z.resize(n + 1), p.resize(m + 1);
    // z函数表示以a_i开头的后缀和a的最长公共前缀
	F(i, 2, n) {
		int r = now + z[now] - 1;
		z[i] = r >= i ? min(z[i - now + 1], r - i + 1) : 0;
		while(a[z[i] + 1] == a[i + z[i]]) z[i] ++;
		if(i + z[i] - 1 > r) now = i;
	} z[1] = n, now = 0;
    // p表示以b_i开头的每一个后缀和a的最长公共前缀
	F(i, 1, m) {
		int r = now + p[now] - 1;
		p[i] = r >= i ? min(z[i - now + 1], r - i + 1) : 0;
		while(i + p[i] <= m && a[p[i] + 1] == b[i + p[i]]) p[i] ++;
		if(i + p[i] - 1 > r) now = i;
	}
}


int n, m, f1[N][N], f2[N][N];
int cal(string s, string t) {
    int n = s.size(), m = t.size();
    auto s2 = s, t2 = t;
    reverse(s2.begin(), s2.end());
    reverse(t2.begin(), t2.end());
    memset(f1, 0, sizeof(f1));
    memset(f2, 0, sizeof(f2));
    F(i, 1, n) {
        vector<int> z, p;
        string q = s.substr(i - 1);
        exkmp(q, t, z, p);
        F(j, i, n) f1[i][j] = min(j - i + 1, z[j - i + 1]);
    }
    F(i, 1, n) {
        vector<int> z, p;
        string q = s2.substr(i - 1);
        exkmp(q, t2, z, p);
        F(j, 1, m) f2[n - i + 1][p[j]] ++;
        Fd(j, n, 0) f2[n - i + 1][j] += f2[n - i + 1][j + 1];
    }   
    int res = 0;
    F(i, 1, n) F(j, i + 2, n) if(s[i - 1] == s[j - 1]) {
        int mn = max(j - i - f1[i][j], 1ll), mx = j - i - 1;
        res += f2[j - 1][mn];
    }
    return res;
}

void sol() {
    string s, t; cin >> s >> t;
    int res = cal(s, t);
    reverse(s.begin(), s.end());
    reverse(t.begin(), t.end());
    res += cal(t, s);
    int n = s.size(), m = t.size();
    F(i, 1, n) {
        vector<int> z, p;
        string q = s.substr(i - 1);
        exkmp(q, t, z, p);
        // F(j, 1, m) cout << p[j] << " \n" [j == m];
        F(j, 1, m) res += p[j];
    }
    cout << res << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    // cin >> t;
    F(i, 1, t) sol();
    return 0;
}
//sldl