Problem:
Score: 200 points

Problem Statement

Today, the memorable AtCoder Beginner Contest 100 takes place. On this occasion, Takahashi would like to give an integer to Ringo.

As the name of the contest is AtCoder Beginner Contest 100, Ringo would be happy if he is given a positive integer that can be divided by 100 exactly D times.

Find the N-th smallest integer that would make Ringo happy.

Constraints

D is 0, 1 or 2.

N is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

D N

Output

Print the N-th smallest integer that can be divided by 100 exactly D times.

Sample Input 1

0 5

Sample Output 1

5

The integers that can be divided by 100 exactly 0 times (that is, not divisible by 100) are as follows: 1, 2, 3, 4, 5, 6, 7, ...

Thus, the 5-th smallest integer that would make Ringo happy is 5.

Sample Input 2

1 11

Sample Output 2

1100

The integers that can be divided by 100 exactly once are as follows: 100, 200, 300, 400, 500, 600, 700, 800, 900, 1 \ 000, 1 \ 100, ...

Thus, the integer we are seeking is 1 \ 100.

Sample Input 3

2 85

Sample Output 3

850000

The integers that can be divided by 100 exactly twice are as follows: 10 \ 000, 20 \ 000, 30 \ 000, ...

Thus, the integer we are seeking is 850 \ 000.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
#define rep(i, j) for (int i = 0; i < j; i++)
using namespace std;
using ll = long long;
using ull = unsigned long long;

int gcd(int x, int y);

const int INF = 1001001001;

/*struct mint {
    ll x; // typedef long long ll;
    mint(ll x = 0) :x((x% mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
    mint operator+(const mint a) const { return mint(*this) += a; }
    mint operator-(const mint a) const { return mint(*this) -= a; }
    mint operator*(const mint a) const { return mint(*this) *= a; }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return *this *= a.inv(); }
    mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream & is, const mint & a) { return is >> a.x; }
ostream& operator<<(ostream & os, const mint & a) { return os << a.x; }*/

int main() {
    ll n, d, ans = 0;
    cin >> d >> n;
    
    ans = n * pow(100, d);
    if (n == 100) {
        ans += 1 * pow(100, d);
    }

    cout << ans << endl;
    return 0;
}

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int gcd(int x, int y) {
    int num[3];
    num[0] = (x > y) ? x : y;
    num[1] = (x <= y) ? x : y;
    num[2] = num[0] % num[1];

    while (num[2]) {
        num[0] = num[1];
        num[1] = num[2];
        num[2] = num[0] % num[1];
    }

    return num[1];
}