#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 10;
const int WIDTH = 4;    // 压4位
const int BASE = 10000; // 基数为10000

int a[N], al;
int b[N], bl;

// 字符串转压位数组（逆序存储）
void str2num(string s, int c[], int &cl) {
  cl = 0;
  for (int i = s.size() - 1; i >= 0; i -= WIDTH) {
    int v = 0;
    int st = max(i - WIDTH + 1, 0); // 最后一组可能不够4个长度
    for (int j = st; j <= i; j++)
      v = v * 10 + (s[j] - '0');
    c[cl++] = v;
  }
}

void mul(int a[], int &al, int b[], int &bl) {
  vector<int> c(N, 0);

  for (int i = 0; i < al; i++)
    for (int j = 0; j < bl; j++)
      c[i + j] += a[i] * b[j];

  al = al + bl + 1;
  for (int i = 0; i < al; i++) {
    c[i + 1] += c[i] / BASE;
    c[i] %= BASE;
  }

  for (int i = 0; i < al; i++)
    a[i] = c[i];

  while (al > 1 && a[al - 1] == 0)
    al--;
}

int main() {
  string x, y;
  cin >> x >> y;

  str2num(x, a, al), str2num(y, b, bl);

  mul(a, al, b, bl);

  for (int i = al - 1; i >= 0; i--)
    printf("%d", a[i]);

  return 0;
}