#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x <<": " << (x) << endl
//在每个函数的入口处执行一次，出口处执行一次。然后就可以快速得知是哪个地方段错误了
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#ifdef LOCAL
#define see(x) cout  << #x << ": " << (x) << endl
#endif
#ifndef LOCAL
#define see(x)
#endif

#define RED     "\033[31m"
#define RESET "\033[0m"
#define alert(x) cerr << RED << x << RESET << endl
#ifndef LOCAL
#define see(x)
#endif

#define Rep(n) for(int _ = 0; _ != (n); ++_)
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define Rng(i, n) for(int i = 0; i != (n); ++i)
#define rng(i, a, b) for(int i = (a); i != (b); ++i)
#define RNG(i, a) for(auto &i: (a))

template<class T> int sign(const T &a) { return a == 0 ? 0 : a > 0 ? 1 : -1; }
template<class T> inline T min(T a, T b, T c){return min(min(a, b), c);}
template<class T> inline T max(T a, T b, T c){return max(max(a, b), c);}
template<class T> void Min(T &a, const T &b){ a = min(a, b); }
template<class T> void Max(T &a, const T &b){ b = max(a, b); }

template<typename T> void println(const T &t) { cout << t << '\n'; }
template<typename T, typename ...Args> void println(const T &t, const Args &...rest) { cout << t << ' '; println(rest...); }

template<typename T> void print(const T &t) { cout << t << ' '; }

template<typename T, typename ...Args> void print(const T &t, const Args &...rest) { cout << t; print(rest...); }

// this overload is chosen when there's only one argument
template<class T> void scan(T &t) { cin >> t; }
template<class T, class ...Args> void scan(T &a, Args &...rest) { cin >> a; scan(rest...); }


int cas;
const double pi = acos(-1);
int mod = 1e9 + 7;

template<class T>
void add_mod(T &a, const T &b) {
    a += b;
    if (a >= mod) a -= mod;
}


using ll = long long;
using ull = unsigned long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pii = pair<int, int>;
using pdd = pair<double, double>;
using pip = pair<int, pii>;
using szt = size_t;
using vi = vector<int>;
using vb = vector<bool>;
using vpii = vector<pii>;

mat operator*(const mat &a, const mat &b) {
    mat c(a.size(), vec(b[0].size()));
    for (int i = 0; i < a.size(); i++) {
        for (int j = 0; j < a[0].size(); j++) {
            if (a[i][j]) { // optimization for sparse matrix
                for (int k = 0; k < b[0].size(); k++) {
                    add_mod(c[i][k], a[i][j] * b[j][k] % mod);
                }
            }
        }
    }
    return c;
}

vec operator*(const mat &a, const vec &b) {
    vec c(a.size());
    for (int i = 0; i < a.size(); i++) {
        for (int j = 0; j < a[0].size(); j++) {
            add_mod(c[i], a[i][j] * b[j] % mod);
        }
    }
    return c;
}

mat pow(mat a, ull n) {
    mat res(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); i++) {
        res[i][i] = 1;
    }
    while (n) {
        if (n & 1) {
            res = res * a;
        }
        a = a * a;
        n >>= 1;
    }
    return res;
}


ll POW(ll x, ll n) {
    ll res = 1;
    for (; n; n /= 2, x *= x, x %= mod) {
        if (n & 1) {
            res *= x;
            res %= mod;
        }
    }
    return res;
}

ll inv(ll x) {
    return POW(x, mod - 2);
}



// 2D rotation
void rotate(double &x, double &y, double theta) {
    double tx = cos(theta) * x - sin(theta) * y;
    double ty = sin(theta) * x + cos(theta) * y;
    x = tx, y = ty;
}

const int BIT_N = 1e5+5;

int bit[BIT_N];

int sum(int x) {
    int res = 0;
    while (x) {
        res += bit[x];
        x -= x & -x;
    }
    return res;
}

int sum(int l, int r) {
    if (l > r) return 0;
    return sum(r) - sum(l - 1);
}

void add(int x, int v, int n) {
    while (x <= n) {
        bit[x] += v;
        x += x & -x;
    }
}



// #include <ext/pb_ds/priority_queue.hpp>

// typedef __gnu_pbds :: priority_queue<pip, less<pip>, __gnu_pbds::thin_heap_tag > Heap;

// Heap h;

// Heap::point_iterator pos[N][N];
const ll INF = LLONG_MAX/10;
const int M = 3000 + 5;






const int N = 305;

priority_queue<pip, vector<pip>, greater<pip>> que;
int dis[N][N], a[N][N];
vpii pos[N * N];
int get_dis(pii a, pii b){
    return abs(a.first - b.first) + abs(a.second - b.second);
}



/// @note 这道题是分层图，题解中提到的暴力DP其实就是DAG上的最短路算法
/// \param p
/// @note 这道题用 dijkstra 算法并不能降低最坏情况复杂度！dijkstra的最坏情况复杂度甚至比暴力DP还高！
void dijkstra(int p) {
    memset(dis, -1, sizeof dis);
    int nxt = a[1][1] == 1 ? 2 : 1;
    RNG (i, pos[nxt]) {
        int t = get_dis(i, {1,1});
        dis[i.first][i.second] = t;
        que.push({t, i});
    }




    while (!que.empty()) {
        auto top = que.top();
        que.pop();
        int x = top.second.first, y = top.second.second;
        int t = dis[x][y];
        if (top.first != t) continue;
        int no = a[x][y];
        if (no == p){
            println(t);
            return;
        }


        RNG (i, pos[no+1]) {
            int &tmp = dis[i.first][i.second], nd = t + get_dis(i, top.second);
            if(tmp == -1 || tmp > nd) {
                tmp = nd;
                que.push({tmp, i});
            }
        }
    }
}
//Clang-Tidy: Prefer transparent functors 'greater<>'



// 被卡IO了
int main() {
    // Single Cut of Failure taut me
    cout << setprecision(10) << std::fixed; // std::fixed 使得所有实数（默认）输出6位小数，即使实际小数位数多于6位。
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
#ifdef LOCAL
    freopen("main.in", "r", stdin);
//    freopen("main.out", "w", stdout);
#endif

    int d, n;
    scan(d, n);

    auto t = POW(100, d);
    println(n==100?101*t:n*t);


    return 0;
}

Measured memory_bytes: 3068