You are given an integer n. You have to apply m operations to it.

In a single operation, you must replace every digit d of the number with the decimal representation of integer d + 1. For example, 1912 becomes 21023 after applying the operation once.

You have to find the length of n after applying m operations. Since the answer can be very large, print it modulo 10^9+7.

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.

The only line of each test case contains two integers n (1 ≤ n ≤ 10^9) and m (1 ≤ m ≤ 2 ⋅ 10^5) — the initial number and the number of operations. 

Output

For each test case output the length of the resulting number modulo 10^9+7.

Example

Input


5
1912 1
5 6
999 1
88 2
12 100


Output


5
2
6
4
2115

Note

For the first test, 1912 becomes 21023 after 1 operation which is of length 5.

For the second test, 5 becomes 21 after 6 operations which is of length 2.

For the third test, 999 becomes 101010 after 1 operation which is of length 6.

For the fourth test, 88 becomes 1010 after 2 operations which is of length 4.
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
// constexpr int MOD = 998244353;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << fixed << setprecision(20);
  }
} iosetup;

template <int M>
struct MInt {
  unsigned int val;
  MInt(): val(0) {}
  MInt(long long x) : val(x >= 0 ? x % M : x % M + M) {}
  static constexpr int get_mod() { return M; }
  static void set_mod(int divisor) { assert(divisor == M); }
  static void init(int x = 10000000) { inv(x, true); fact(x); fact_inv(x); }
  static MInt inv(int x, bool init = false) {
    // assert(0 <= x && x < M && std::__gcd(x, M) == 1);
    static std::vector<MInt> inverse{0, 1};
    int prev = inverse.size();
    if (init && x >= prev) {
      // "x!" and "M" must be disjoint.
      inverse.resize(x + 1);
      for (int i = prev; i <= x; ++i) inverse[i] = -inverse[M % i] * (M / i);
    }
    if (x < inverse.size()) return inverse[x];
    unsigned int a = x, b = M; int u = 1, v = 0;
    while (b) {
      unsigned int q = a / b;
      std::swap(a -= q * b, b);
      std::swap(u -= q * v, v);
    }
    return u;
  }
  static MInt fact(int x) {
    static std::vector<MInt> f{1};
    int prev = f.size();
    if (x >= prev) {
      f.resize(x + 1);
      for (int i = prev; i <= x; ++i) f[i] = f[i - 1] * i;
    }
    return f[x];
  }
  static MInt fact_inv(int x) {
    static std::vector<MInt> finv{1};
    int prev = finv.size();
    if (x >= prev) {
      finv.resize(x + 1);
      finv[x] = inv(fact(x).val);
      for (int i = x; i > prev; --i) finv[i - 1] = finv[i] * i;
    }
    return finv[x];
  }
  static MInt nCk(int n, int k) {
    if (n < 0 || n < k || k < 0) return 0;
    if (n - k > k) k = n - k;
    return fact(n) * fact_inv(k) * fact_inv(n - k);
  }
  static MInt nPk(int n, int k) { return n < 0 || n < k || k < 0 ? 0 : fact(n) * fact_inv(n - k); }
  static MInt nHk(int n, int k) { return n < 0 || k < 0 ? 0 : (k == 0 ? 1 : nCk(n + k - 1, k)); }
  static MInt large_nCk(long long n, int k) {
    if (n < 0 || n < k || k < 0) return 0;
    inv(k, true);
    MInt res = 1;
    for (int i = 1; i <= k; ++i) res *= inv(i) * n--;
    return res;
  }
  MInt pow(long long exponent) const {
    MInt tmp = *this, res = 1;
    while (exponent > 0) {
      if (exponent & 1) res *= tmp;
      tmp *= tmp;
      exponent >>= 1;
    }
    return res;
  }
  MInt &operator+=(const MInt &x) { if((val += x.val) >= M) val -= M; return *this; }
  MInt &operator-=(const MInt &x) { if((val += M - x.val) >= M) val -= M; return *this; }
  MInt &operator*=(const MInt &x) { val = static_cast<unsigned long long>(val) * x.val % M; return *this; }
  MInt &operator/=(const MInt &x) { return *this *= inv(x.val); }
  bool operator==(const MInt &x) const { return val == x.val; }
  bool operator!=(const MInt &x) const { return val != x.val; }
  bool operator<(const MInt &x) const { return val < x.val; }
  bool operator<=(const MInt &x) const { return val <= x.val; }
  bool operator>(const MInt &x) const { return val > x.val; }
  bool operator>=(const MInt &x) const { return val >= x.val; }
  MInt &operator++() { if (++val == M) val = 0; return *this; }
  MInt operator++(int) { MInt res = *this; ++*this; return res; }
  MInt &operator--() { val = (val == 0 ? M : val) - 1; return *this; }
  MInt operator--(int) { MInt res = *this; --*this; return res; }
  MInt operator+() const { return *this; }
  MInt operator-() const { return MInt(val ? M - val : 0); }
  MInt operator+(const MInt &x) const { return MInt(*this) += x; }
  MInt operator-(const MInt &x) const { return MInt(*this) -= x; }
  MInt operator*(const MInt &x) const { return MInt(*this) *= x; }
  MInt operator/(const MInt &x) const { return MInt(*this) /= x; }
  friend std::ostream &operator<<(std::ostream &os, const MInt &x) { return os << x.val; }
  friend std::istream &operator>>(std::istream &is, MInt &x) { long long val; is >> val; x = MInt(val); return is; }
};
namespace std { template <int M> MInt<M> abs(const MInt<M> &x) { return x; } }
using ModInt = MInt<MOD>;

constexpr int M = 200000;
ModInt dp[10][M + 1]{};

void solve() {
  string n; int m; cin >> n >> m;
  ModInt ans = 0;
  for (char c : n) ans += dp[c - '0'][m];
  cout << ans << '\n';
}

int main() {
  REP(i, 10) dp[i][0] = 1;
  FOR(j, 1, M + 1) REP(i, 10) {
    for (char c : to_string(i + 1)) dp[i][j] += dp[c - '0'][j - 1];
  }
  int t; cin >> t;
  while (t--) solve();
  return 0;
}
