#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
#define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++)
#define REP(i, n) for (long long i = 1; i < (long long)(n); i++)
typedef long long ll;
#define updiv(N,X) (N + X - 1) / X
#define l(n) n.begin(),n.end()
#define YesNo(Q) Q==1?cout<<"Yes":cout<<"No"
using P = pair<int, int>;
using mint = modint;
const int MOD = 998244353LL;
const ll INF = 999999999999LL;
vector<long long> fact, fact_inv, inv;
/*  init_nCk :二項係数のための前処理
    計算量:O(n)
*/
template <typename T>
void input(vector<T> &v){
 rep(i,v.size()){cin>>v[i];}
  return;
}
void init_nCk(int SIZE) {
    fact.resize(SIZE + 5);
    fact_inv.resize(SIZE + 5);
    inv.resize(SIZE + 5);
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < SIZE + 5; i++) {
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;
    }
}
/*  nCk :MODでの二項係数を求める(前処理 int_nCk が必要)
    計算量:O(1)
*/
long long nCk(int n, int k) {
    assert(!(n < k));
    assert(!(n < 0 || k < 0));
    return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

ll POW(ll a,ll n){
  long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}

mint pro(ll a){
  if(a<=1){return (mint)(0);}
  else{
    mint x = (a-1)/2;
    mint y = a/2;
    mint ret = (x*x+x+(y-1)*y);
    ret /= 2;
    return ret;
  }
}
mint pro2(ll a,ll b){
   mint r1 = (b-1+(b-a));
   r1 *= a;
   r1 /= 2;
   if((a-1)%2==0){
     r1 -= (a+1)/2; 
   }
   else{
     r1 -= a/2;
   }
   r1 /= 2;
   return r1;
}
void test(){
  ll n,k;cin>>n>>k;
  mint ret = pro(POW(10LL,k));
  //mint ret = 1;
  ret *= pro(n/POW(10LL,k));
  mint ret2 = (n/POW(10LL,k))/2;
  ret2 *= pro2(n%POW(10LL,k),POW(10LL,k));
  cout << (ret2+ret).val() << endl;
  
}
int main(){
  //cout << pro2(1,10).val() << endl;
  int t;cin>>t;
  while(t--){
    test();
  }
}