Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.

You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.

Input

The first line contains t (1 ≤ t ≤ 100) — the number of test cases.

Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).

Output

Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.

Example

Input


3
1 11
4 2
191 31415926


Output


1
0
1337

Note

There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
#include <bits/stdc++.h>
const long long int mod = 9999999999999983;
long long int primes[6] = {1125899906842597, 1495921043, 1005985879,
                           1495921043,       1005985879, 1495921043};
using namespace std;
vector<long long int> adj[2000009];
long long int parent[2000009];
long long int vis[2000009];
long long int level[2000009];
long long int dist[2000009];
long long int dp[100009];
long long int brr[2000009];
long long int crr[2000009];
long long int hashing[2000009];
long long int ar[509][509];
long long int br[509][509];
long long int cr[509][509];
long long int multiply(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int sub(long long int a, long long int b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
int main() {
  int start_s = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  vector<long long int> vect, vt, vct;
  vector<long long int>::iterator itt;
  set<long long int> st;
  set<long long int>::iterator itr;
  map<long long int, long long int> mp, mpp, mp1, mp2;
  map<long long int, long long int>::iterator it;
  long long int i, j, k, n, m, res1, res2, q, t, a, b, c,
      maxi = INT_MIN, mini = INT_MAX, sum = 0, ans = 0, res = 0, val = 0,
      ans1 = 0, ans2 = 0, rem = 0, diff = 0, cnt = 0, l, r, flag = 0, e, index,
      val1 = 0, val2 = 0, z, h = 0, u, v, mid, len, tot, fl = 0;
  string str, str1, str2;
  cin >> t;
  long long int arr[] = {9,      99,      999,      9999,      99999,
                         999999, 9999999, 99999999, 999999999, 9999999999};
  while (t--) {
    cin >> a >> b;
    cnt = 0;
    for (i = 0; i < 10; i++) {
      if (arr[i] <= b) {
        cnt++;
      } else
        break;
    }
    cout << a * cnt << "\n";
  }
}
