/*
You have a list in which you store strings. The list is indexed from 1. Initially, the list contains only one string \(S\).

There are three different types of tasks that you are asked to perform:

\(1 \quad p \quad n \quad c\): Create a new string of length \(p\) using the \(n^{th}\) string from the list such that all characters from \(1\) to \(p-1\) in the new string are the same as in the old string and the \(p^{th}\) character is \(c\). Append this string to the end of the list.

\(2 \quad p \quad n\): Create a new string of length \(p\) using the \(n^{th}\) string from the list such that all the characters from \(1\) to \(p\) in the new string are the same as in the old string. Append this string to the end of the list.

\(3 \quad l \quad r \quad s\): Print \(yes\) if string \(s\) appears as the prefix of any string in the range \([l,r]\) of the list, else print \(no\).

Input format

First line: String \(S\) represents the only string that is initially in the list
Next line: \(Q\) represents the number of queries to be performed
Next \(Q\) lines: Each line describes one of the three types of tasks that can be performed. All the tasks are valid.
Output format

For every task of type \(3\), print \(yes\) or \(no\).

Constraints

\(|S| \le 10^5\)

\(Q \le 3*10^{5}\)

Constraints that apply to all the queries

For the query to perform the task of type \(1\): \(0 \le p-1 \le | size \; of \; n^{th} \; string|\)

For the query to perform the task \(2\): \(1 \le p \lt | size \; of \; n^{th} \; string|\)

\(1 \le n \le |current \; size \; of \; list|\)

\(c \in [a,b]\), i.e. all the characters are lowercase English alphabets

\(1 \le l \le r \le |current \quad size \quad of \quad list|\)

\(\sum_{Q} |s_i| \le 10^6\)

\(s_{i} \ne s_{j}\) if \(i \ne j\), for all the strings in the list. i.e. all the strings in the list are unique

\(c \in [a, z]\ (not [a, b])\)
*/

#include <bits/stdc++.h>
using namespace std;
#define nl "\n"
#define fastio          std::ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define LL              long long
#define LD              long double
// #define MAX             200005
#define MOD             1000000007
#define REP(i,n)        for(__typeof(n) i=0; i<n; ++i)
#define CREP(i,n)       for(__typeof(n) i=n-1; i>=0; --i)
#define MYREP(i,a,b)    for(__typeof(a) i=a; i<=b; ++i)
#define MYCREP(i,a,b)   for(__typeof(a) i=b; i>=a; --i)
#define SET(a, b)       memset(a, b, sizeof(a))
#define pb              push_back
#define mp              make_pair
#define fi              first
#define sec             second
#define aLL(c)          (c).begin(),(c).end()
#define aLLr(c)         (c).rbegin(),(c).rend()
#define loop(c,i)       for(typeof(c.begin()) i = c.begin(); i != c.end(); i++)
#define present(c,x)    ((c).find(x) != (c).end())
#define cpresent(c,x)   (find(aLL(c),x) != (c).end())
#define BLOCK 555   // Optimum in Most Cases    Mo's Algo see Anudeep Bookamrks

long mul(long a, long b, long c) {long long res = (long long)a * b; return (res >= c ? res % c : res);}

template <typename T> T mod(T a, T b) {return (a < b ? a : a % b);}
//(a - b) % c
template<typename T> T mod_neg(T a, T b) {a = mod(a, b); if (a < 0) {a += b;} return a;}
//(a * b) % c for very large numbers, to avoid overflows
//O(1) hack taken from http://codeforces.com/blog/entry/15884
LL mulmod(LL a, LL b, LL m) {LL q = (LL)(((LD)a * (LD)b) / (LD)m); LL r = a * b - q * m; if (r > m)r %= m; if (r < 0)r += m; return r;}
//(a ^ n)
//Complexity: O(log (n))
template <typename T>T expo(T e, T n) {T x = 1, p = e; while (n) {if (n & 1)x = x * p; p = p * p; n >>= 1;} return x;}
//(a ^ n) % m
//Complexity : O(log (n))

template <typename T>T power(T e, T n, T m) {T x = 1, p = e; while (n) {if (n & 1)x = mod(x * p, m); p = mod(p * p, m); n >>= 1;} return x;}
template <typename T>T powerL(T e, T n, T m) {T x = 1, p = e; while (n) {if (n & 1)x = mulmod(x, p, m); p = mulmod(p, p, m); n >>= 1;} return x;}

template <typename T> T InverseEuler(T a, T m) {return (a == 1 ? 1 : power(a, m - 2, m));}
//gcd of 2 numbers a, b
//Complexity: O(log(max(a, b)))
template<typename T> T gcd(T a, T b) { return (b ? __gcd(a, b) : a); }
//lcm of 2 numbers a and b
//Complexity: O(log(max(a, b)))
template <typename T> T lcm(T a, T b) {return (a * (b / gcd(a, b)));}
template<typename T> T extended_euclid(T a, T b, T &x, T &y) {T xx = 0, yy = 1; y = 0; x = 1; while (b) {T q = a / b, t = b; b = a % b; a = t; t = xx; xx = x - q * xx; x = t; t = yy; yy = y - q * yy; y = t;} return a;}
template<typename T> T mod_inverse(T a, T n) {T x, y; T d = extended_euclid(a, n, x, y); return (d > 1 ? -1 : mod_neg(x, n));}
template<class T> T fast_gcd(T u, T v) {
    long shl = 0; while ( u && v && u != v) {
        T eu = u & 1; u >>= eu ^ 1; T ev = v & 1; v >>= ev ^ 1;
        shl += (~(eu | ev) & 1); T d = u & v & 1 ? (u + v) >> 1 : 0; T dif = (u - v) >> (sizeof(T) * 8 - 1); u -= d & ~dif; v -= d & dif;
    } return std::max(u, v) << shl;
}

// Most immportant one best optimised
template<class T> bool isPrime(T x) {
    if (x <= 4 || x % 2 == 0 || x % 3 == 0) return x == 2 || x == 3;
    for (T i = 5; i * i <= x; i += 6) if (x % i == 0 || x % (i + 2) == 0) return 0; return 1;
}

/*

#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
// tree defined...
using namespace __gnu_pbds;
typedef tree <long, null_type, less_equal<long>, rb_tree_tag,
             tree_order_statistics_node_update> ordered_set;
//All indexing is 0-based
//methods: find_by_order(k); & order_of_key(k);
//To make it an ordered_multiset, use pairs of (value, time_of_insertion)
//to distinguish values which are similar.

typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
             tree_order_statistics_node_update> ordered_multiset;

template<class key, class value, class cmp = std::less<key>>
using tordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
//ordered_map<int, int> my_map;
*/

/*
const long MAXS = 1e6 + 5;
LL bit[MAXS];       //1-based indexing for bit
long n;
//Complexity: O(log n)
void update(long idx, long val) {
    while (idx <= MAXS) {
        bit[idx] += val;
        idx += (idx & -idx);
    }
}

//Complexity: O(log n)
long long query(long idx) {
    long sum = 0;
    while(idx > 0) {
        sum += bit[idx];
        idx -= (idx & -idx);
    }
    return sum;
}
*/

const int N = 1e6 + 5;
int seive[N];

void pre() {
    fill(seive + 2, seive + N, 1);

    for (int i = 2; i < N; i++) {
        if (seive[i]) {
            for (int j = 2; j * i <= N; j++) {
                seive[i * j] = 0;
            }
        }
    }
}

int main(){

    // ༼ つ ◕_◕ ༽つ

    int n = 3;
    int a[3][3];

    REP(i,3){
        REP(j,3){
            cin>>a[i][j];
        }
    }

    for(int a1 = 0; a1 <= 100; a1++){
        for(int a2 = 0; a2 <= 100; a2++){
            for(int a3 = 0;a3 <= 100; a3++){
                
                int b1, b2, b3, fl1 = 0, fl2 = 0, fl3 = 0;
                b1 = a[0][0] - a1;
                if(b1 == a[1][0] - a2 && b1 == a[2][0] - a3){
                    fl1 = 1;
                }
                b2 = a[0][1] - a1;
                if(b2 == a[1][1] - a2 && b2 == a[2][1] - a3){
                    fl2 = 1;
                }
                b3 = a[0][2] - a1;
                if(b3 == a[1][2] - a2 && b3 == a[2][2] - a3){
                    fl3 = 1;
                }
                if(fl1 && fl2 && fl3){
                    cout<<"Yes";
                    return 0;
                }
            }
        }
    }
    cout<<"No";

}

Measured memory_bytes: 256