#include<set>

#include<map>

#include<stack>

#include<queue>

#include<cmath>

#include<cstdio>

#include<cctype>

#include<string>

#include<vector>

#include<climits>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define max(a, b) (a > b ? a : b)

#define min(a, b) (a < b ? a : b)

#define mst(a) memset(a, 0, sizeof(a))

#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")

using namespace std;

typedef long long ll;

typedef pair<int, int> P;

const double eps = 1e-;

const int INF = 0x3f3f3f3f;

const ll ll_INF = ;

const int maxn = 1e3+;

char num1[maxn], num2[maxn], res[maxn];

void sum() { //大整数加法

    mst(res);

    int len1 = strlen(num1);

    int len2 = strlen(num2);

    reverse(num1, num1+len1); //反转第一个数

    reverse(num2, num2+len2); //反转第二个数

    int kase = ;

    while(kase < len1 && kase < len2) { //重叠部分相加

        res[kase+] = (res[kase] + num1[kase] + num2[kase] - ''*)/; //模拟进位相加

        res[kase] = (res[kase] + (num1[kase] + num2[kase] - ''*))% + '';

        ++kase;

    }

    while(kase < len1) { //非重叠部分相加

        res[kase+] = (res[kase] + num1[kase] - '')/;

        res[kase] = (res[kase] + num1[kase] - '')% + '';

        ++kase;

    }

    while(kase < len2) { //同上

        res[kase+] = (res[kase] + num2[kase] - '')/;

        res[kase] = (res[kase] + num2[kase] - '')% + '';

        ++kase;

    }

    if (res[kase]) { //判断最后一次相加有没有进位

        res[kase] += '';

        res[++kase] = '\0';

    }

    else

        res[kase] = '\0';

}

bool okk() { //判断回文

    int len = strlen(res);

    for (int i = ; i<len/; ++i)

        if (res[i] != res[len-i-])

            return false;

    return true;

}

int main(void) {

    while(~scanf("%[^\n]%*c", num1)) {

        bool ok = false;

        strcpy(num2, num1);

        for (int i = ; i<; ++i) {

            reverse(num2, num2+strlen(num2));

            sum(); //将反转过后的数与原数字相加

            if (okk()) { //判断是否满足回文

                ok = true;

                break;

            }

            strcpy(num1, res); //将相加后的结果复制给另外两个数字准备下次计算

            strcpy(num2, res);

        }

        printf(ok ? "%s\n" : "NO\n", res);

    }

    return ;

}