THẦN SỐ HỌC

#define task "NROLOGY"
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

constexpr int N = 5e2 + 5;
constexpr ll mod = 1e9 + 7;
string b, s;

ll dp[N][N][2][3], d[N][N];

int n, m;

void Read()
{
    cin >> s >> b;

    reverse(b.begin(), b.end());

    while (b.size() < s.size())
        b.push_back('0');
    reverse(b.begin(), b.end());

    n = b.size();
    m = s.size();
    b = " " + b + " ";
    s = " " + s + " ";
}

ll toll(const string s)
{
    ll ans(0);

    for (auto i : s)
        ans = ans * 10 + i - '0';

    return ans;
}

ll Pow(ll a, ll b) {
    ll ans(1);
    for(; b; b >>= 1) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod;
    }
    return ans;
}

ll f(int l, int r, bool okl, int okr)
{

    ll &ans = dp[l][r][okl][okr];

    if (ans != -1)
        return ans;

    int pos = n - (l + (n + 1 - r));

    if (pos == 0)
        return ans = (okl || okr != 2);

    ans = 0;

    if (okl || s[pos] <= b[l + 1])
        (ans += f(l + 1, r, okl || s[pos] < b[l + 1], okr)) %= mod;

    {
        int t;

        if (s[pos] > b[r - 1])
            t = 2;
        else if (s[pos] < b[r - 1])
            t = 1;
        else
            t = okr;

        (ans += f(l, r - 1, okl, t)) %= mod;
    }

    return ans;
}

void Solve()
{

    if (s.size() < b.size())
        cout << Pow(2, m - 1);
    else
    {
        memset(dp, -1, sizeof dp);
        cout << f(0, n + 1, 0, 0) * 500000004 % mod; // 500000004 = 2^(mod - 2)
    }
}

int32_t main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    if (fopen(task ".INP", "r"))
    {
        freopen(task ".INP", "r", stdin);
        freopen(task ".OUT", "w", stdout);
    }
    Read();
    Solve();
}