Toshizo is the manager of a convenience store chain in Hakodate. Every day, each of the stores in his chain sends him a table of the products that they have sold. Toshizo's job is to compile these figures and calculate how much the stores have sold in total.

The type of a table that is sent to Toshizo by the stores looks like this (all numbers represent units of products sold):



Store1      Store2      Store3         Totals
Product A                 -5          40          70      |    105
Product B                 20          50          80      |    150
Product C                 30          60          90      |    180
-----------------------------------------------------------
Store's total sales       45         150         240           435


By looking at the table, Toshizo can tell at a glance which goods are selling well or selling badly, and which stores are paying well and which stores are too empty. Sometimes, customers will bring products back, so numbers in a table can be negative as well as positive.

Toshizo reports these figures to his boss in Tokyo, and together they sometimes decide to close stores that are not doing well and sometimes decide to open new stores in places that they think will be profitable. So, the total number of stores managed by Toshizo is not fixed. Also, they often decide to discontinue selling some products that are not popular, as well as deciding to stock new items that are likely to be popular. So, the number of products that Toshizo has to monitor is also not fixed.

One New Year, it is very cold in Hakodate. A water pipe bursts in Toshizo's office and floods his desk. When Toshizo comes to work, he finds that the latest sales table is not legible. He can make out some of the figures, but not all of them. He wants to call his boss in Tokyo to tell him that the figures will be late, but he knows that his boss will expect him to reconstruct the table if at all possible.

Waiting until the next day won't help, because it is the New Year, and his shops will be on holiday. So, Toshizo decides either to work out the values for himself, or to be sure that there really is no unique solution. Only then can he call his boss and tell him what has happened.

But Toshizo also wants to be sure that a problem like this never happens again. So he decides to write a computer program that all the managers in his company can use if some of the data goes missing from their sales tables. For instance, if they have a table like:



Store 1     Store 2    Store 3            Totals
Product A              ?           ?         70        |     105
Product B              ?          50          ?        |     150
Product C             30          60         90        |     180
--------------------------------------------------------
Store's total sales   45         150        240              435


then Toshizo's program will be able to tell them the correct figures to replace the question marks. In some cases, however, even his program will not be able to replace all the question marks, in general. For instance, if a table like:


Store 1     Store 2            Totals
Product A              ?           ?        |      40
Product B              ?           ?        |      40
---------------------------------------------
Store's total sales   40          40               80


is given, there are infinitely many possible solutions. In this sort of case, his program will just say "NO". Toshizo's program works for any data where the totals row and column are still intact. Can you reproduce Toshizo's program?



Input

The input consists of multiple data sets, each in the following format:


p s
row1
row2
...
rowp
totals


The first line consists of two integers p and s, representing the numbers of products and stores, respectively. The former is less than 100 and the latter is less than 10.

They are separated by a blank character. Each of the subsequent lines represents a row of the table and consists of s+1 elements, each of which is either an integer or a question mark. The i-th line (1 <= i <= p) corresponds to the i-th product and the last line the totals row. The first s elements in a line represent the sales of each store and the last one the total sales of the product. Numbers or question marks in a line are separated by a blank character. There are no missing numbers in the totals column or row. There is at least one question mark in each data set.

The known numbers in the table except the totals row and column is between -1,000 and 1,000, inclusive. However, unknown numbers, represented by question marks, may not be in this range. The total sales number of each product is between -10,000 and 10,000, inclusive. The total sales number of each store is between -100,000 and 100,000, inclusive.

There is a blank line between each data set in the input, and the input is terminated by a line consisting of a zero.

Output

When there is a unique solution, your program should print the missing numbers in the occurrence order in the input data set. Otherwise, your program should print just "NO". The answers of consecutive data sets should be separated by an empty line. Each non-empty output line contains only a single number or "NO".

Example

Input

3 3
? ? 70 105
? 50 ? 150
30 60 90 180
45 150 240 435

2 2
? ? 40
? ? 40
40 40 80

2 3
? 30 40 90
50 60 70 180
70 90 110 270

0


Output

-5
40
20
80

NO

20
#include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int main()
{
    bool first = true;

    for(;;){
        int p, s;
        cin >> p >> s;
        if(p == 0)
            return 0;
        ++ p;
        ++ s;

        vector<vector<int> > num(p, vector<int>(s, 0));
        vector<vector<bool> > missing(p, vector<bool>(s, false));
        vector<int> missingNum(p+s, 0);
        for(int i=0; i<p; ++i){
            for(int j=0; j<s; ++j){
                if(!(cin >> num[i][j])){
                    cin.clear();
                    cin.ignore();
                    missing[i][j] = true;
                    ++ missingNum[i];
                    ++ missingNum[p+j];
                }
                if((i == p-1) ^ (j == s-1))
                    num[i][j] *= -1;
            }
        }

        vector<vector<bool> > check = missing;
        queue<int> q;
        for(int i=0; i<p+s; ++i){
            if(missingNum[i] == 1)
                q.push(i);
        }
        while(!q.empty()){
            int y = q.front();
            int x = -1;
            q.pop();

            int tmp = 0;
            if(y < p){
                for(int i=0; i<s; ++i){
                    tmp -= num[y][i];
                    if(check[y][i])
                        x = i;
                }
            }else{
                x = y - p;
                y = -1;
                for(int i=0; i<p; ++i){
                    tmp -= num[i][x];
                    if(check[i][x])
                        y = i;
                }
            }
            if(y == -1 || x == -1)
                continue;

            num[y][x] = tmp;
            check[y][x] = false;
            -- missingNum[y];
            -- missingNum[p+x];
            if(missingNum[y] == 1)
                q.push(y);
            if(missingNum[p+x] == 1)
                q.push(p+x);
        }

        if(first)
            first = false;
        else
            cout << endl;

        if(*max_element(missingNum.begin(), missingNum.end()) > 0){
            cout << "NO" << endl;
        }else{
            for(int i=0; i<p; ++i){
                for(int j=0; j<s; ++j){
                    if(missing[i][j])
                        cout << num[i][j] << endl;
                }
            }
        }
    }
}