#include <bits/stdc++.h>
//#include <iostream>
#include <atcoder/all>
#include <math.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rrep(i, n) for(int i = (n-1); i >= 0; i--)
using namespace std;
using namespace atcoder;
typedef long long ll;
const int MOD = 1000000007;
//const int MOD =  998244353;
 
const ll INF = 1LL<<61;
const int IINF = 1000000000;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

typedef pair<int, int> PI;
typedef pair<ll, ll> P;
typedef pair<string, ll> PS;

using mint = modint1000000007;

#include <regex>
int main(){

    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M; cin >> N >> M;
    string op; cin >> op;
    vector<ll> A(N), B(M);
    rep(i,M) cin >> B[i];
    rep(i,N) cin >> A[i];

    vector<vector<ll>> ans(N, vector<ll>(M,0));
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            if(op == "+") ans[i][j] = A[i]+B[j];
            else ans[i][j] = A[i] * B[j];
        }
    }
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            printf("%Ld%c", ans[i][j], j==M-1?'\n':' ');
        }
    }
}