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

You are given two integers AA and BB, calculate the number of pairs (a,b)(a,b) such that 1≤a≤A1≤a≤A, 1≤b≤B1≤b≤B, and the equation a?b+a+b=conc(a,b)a?b+a+b=conc(a,b) is true; conc(a,b)conc(a,b) is the concatenation of aa and bb (for example, conc(12,23)=1223conc(12,23)=1223, conc(100,11)=10011conc(100,11)=10011). aa and bb should not contain leading zeroes.

Input
The first line contains tt (1≤t≤1001≤t≤100) — the number of test cases.

Each test case contains two integers AA and BB (1≤A,B≤109)(1≤A,B≤109).

Output
Print one integer — the number of pairs (a,b)(a,b) such that 1≤a≤A1≤a≤A, 1≤b≤B1≤b≤B, and the equation a?b+a+b=conc(a,b)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=1a=1, b=9b=9 (1+9+1?9=191+9+1?9=19).
對照樣例進行打表找規律。
規律如下：

代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;ll A,B; ll C[]={9,99,999,9999,99999,999999,9999999,99999999,999999999}; //一共是9個，一開始少了一個，wa了兩次 int main() {int t;scanf("%d",&t);while(t--){cin>>A>>B;int pos=upper_bound(C,C+9,B)-C;cout<<A*(ll)pos<<endl;}return 0; }

努力加油a啊，(^o)/~

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