#include<bits/stdc++.h>
#define ll long long
#define pii make_pair
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,b,a) for(int i=b;i>=a;--i)
const ll inf = 1145141919810;
using namespace std;
inline ll read() {
	ll x=0,f=1;
	char c=getchar();
	while (c<'0' || c>'9') {
		if (c=='-')  f=-1;
		c=getchar();
	}
	while (c>='0' && c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
inline void print(ll x) {
	if(x < 0) putchar('-'), x = -x;
	if(x > 9) print(x / 10);
	putchar(x % 10 + '0');
	return ;
}
inline void pprint(ll x) {
	print(x);
	puts("");
}

const int N = 5010;
ll f[N][N];
ll g[N][N];
int lcp[N][N];
int n, m;
char s[N], t[N];
void init() {
	scanf("%s", s + 1);
	scanf("%s", t + 1);
	n = strlen(s + 1);
	m = strlen(t + 1);
	for(int i = n; i >= 1; --i) {
		for(int j = m; j >= 1; --j) {
			if(s[i] == t[j]) {
				lcp[i][j] = lcp[i + 1][j + 1] + 1;
			}
		}
	}
	
	rep(i, 1, n) { //统计lcp的数目
		rep(j, 1, m) {
			f[i][lcp[i][j]]++;//从i开始有一个lcp是lcp[i][j]的lcp
		}
	}
	rep(i, 1, n) {
		per(j, 5000, 0) {
			f[i][j] += f[i][j + 1];
		}
	}
	per(i, n, 1) {
		rep(j, 1, 5000) {
			f[i][j] += f[i + 1][j - 1];
		}
	}

	rep(j, 1, m) {
		rep(i, 1, n) {
			g[j][lcp[i][j]]++;
		}
	}
	rep(j, 1, m) {
		per(i, 5000, 0) {
			g[j][i] += g[j][i + 1];
		}
	}
	per(j, m, 1) {
		rep(i, 1, 5000) {
			g[j][i] += g[j + 1][i - 1];
		}
	}
}


void slv() {
	ll ans = 0;
	rep(i, 1, n) rep(j, 1, m) ans += lcp[i][j];
	
	memset(lcp, 0, sizeof lcp);
	per(i, n, 1) {
		per(j, n, 1) {
			if(s[i] == s[j]) {
				lcp[i][j] = lcp[i + 1][j + 1] + 1;
			}
		}
	}

	rep(i, 1, n) {
		rep(j, i + 2, n) {//i j是两个端点
			int r = min(j, i + lcp[i][j] + 1);//r是禁止端点
			int len = j - i - 1;
			ans += f[i + 1][len] - f[r][len - (r - (i + 1))];
		}
	}
	lcp[m + 1][m + 1] = 0;
	
	memset(lcp, 0, sizeof lcp);
	per(i, m, 1){
		per(j, m, 1){
			if(t[i] == t[j]){
				lcp[i][j] = lcp[i + 1][j + 1] + 1;
			}
		}
	}
	rep(i, 1, m) {
		rep(j, i + 2, m) {//i j是两个端点
			int r = min(j, i + lcp[i][j] + 1);//r是禁止端点
			int len = j - i - 1;
			ans += g[i + 1][len] - g[r][len - (r - (i + 1))];
		}
	}
	pprint(ans);
}
int main() {
	init();
	slv();
	return 0;
}