#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<vector>
using namespace std;
const int N = 1e6 + 10;

vector<int> add(vector<int>& A, vector<int>& B)
{
    vector<int> C;

    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i++)
    {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);//相加后的某一位的数字
        t /= 10;//进位操作
    }
    if (t) C.push_back(1);//如果t不为零最高位进1
    return C;
}//大数加法
int main()
{
    string a, b;
    vector<int> A, B;

    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');//逆序存放便于计算

    auto C = add(A, B);
    for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
    return 0;
}

#include<iostream>
#include<vector>
using namespace std;
int main()
{
	string a;
	vector<int> A;
	cin >> a;
	for (int i = 0; i < a.size(); i++)
	{
		A.push_back(a[i] - '0');
	}
	int l = 0;
	int r = a.size()-1;
	while (l < r)
	{
		if (a[l] == a[r])
		{
			l++;
			r--;
		}
		else
		{
			printf("It is not a palindrome!");
			break;
		}
	}
	if (l == r)
	{
		printf("It is a palindrome!");
	}

	return 0;
}//判断回文数