For a positive integer n, we denote the integer obtained by reversing the decimal notation of n (without leading zeroes) by rev(n). For example, rev(123) = 321 and rev(4000) = 4.

You are given a positive integer D. How many positive integers N satisfy rev(N) = N + D?

Constraints

* D is an integer.
* 1 ≤ D < 10^9

Input

Input is given from Standard Input in the following format:


D


Output

Print the number of the positive integers N such that rev(N) = N + D.

Examples

Input

63


Output

2


Input

75


Output

0


Input

864197532


Output

1920
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
ll Power[20];
ll dfs(ll l , ll r , ll d) {
    if(l >= r) {
        if(d == 0) return l == r ? 10 : 1;
        return 0;
    }
    if(d % 10 == 0) {
        if(l != 1) return 10 * dfs(l + 1 , r - 1 , d / 10);
        return 9 * dfs(l + 1 , r - 1 , d / 10);
    }
    ll differ = d % 10 - 10;
    ll next = d - differ + differ * Power[r - l];
    if(l == 1) return (9 + differ) * dfs(l + 1 , r - 1 , abs(next) / 10);
    return (10 + differ) * dfs(l + 1 , r - 1 , abs(next) / 10);
}
int main() {
    ll d;
    cin >> d;
    Power[0] = 1;
    for(int i = 1 ; i <= 18 ; i++) {
        Power[i] = Power[i - 1] * 10;
    }
    ll ans = 0;
    for(int i = 2 ; i <= 18 ; i++) {
        ans += dfs(1 , i , d);
    }
    cout << ans << endl;
    return 0;
}