#include <iostream>
using namespace std;
int n;//数的长度有在加的过程中变化 bool isPalindrome(int a[]){//判断是否是回文数 int mid = 0;if(n%2 == 0){mid = n/2-1;for(int i = 0, j = n-1; i <= mid && j >= mid+1; i++, j--)if(a[i] != a[j])return 0;} else{mid = n/2;for(int i = 0, j = n-1; i < mid && j > mid; i++, j--)if(a[i] != a[j])return 0;} return 1;
}void adverse(int a[], int adv[]){//倒序 int k = 0;for(int i = n-1; i >= 0; i--)adv[k++] = a[i];
}void add(int a[], int adv[]){//大数相加 int temp1[1010] = {0};int temp2[1010] = {0};int i = 0;while(i < n || a[i] != 0){ int up = (a[i]+adv[i])/10;int now = (a[i]+adv[i])%10;temp1[i++] = now;a[i] += up;}n = i;adverse(temp1, temp2);for(int j = 0; j < i; j++)a[j] = temp2[j];
}int main(){string b;cin >> b;int a[1005] = {0};int adv[1005] = {0};for(int i = 0; i < b.length(); i++)a[i] = b[i] - '0';n = b.length();int count = 0;while(!isPalindrome(a)) {count++; if(count == 11)break;adverse(a, adv);for(int i = 0; i < n; i++)cout << a[i];cout << " + ";for(int i = 0; i < n; i++)cout << adv[i];add(a, adv);cout << " = "; for(int i = 0; i < n; i++)cout << a[i];cout << endl; }if(count == 11)cout << "Not found in 10 iterations." << endl;else{for(int i = 0; i < n; i++)cout << a[i];cout << " is a palindromic number." << endl;} return 0;
}