【Gym - 101498H Palindrome Number】 思维

H - Palindrome Number

A palindrome number is a number that can be read the same way from left to right and from right to left. For example: 961169, 111, 554455 are palindromes, while 856, 10180, 7226 are not.

Find the largest palindrome number that satisfies the following rules:

The number of its digits is equal to n, and must not contain leading zeros.
The sum of its digits is equal to s.
If there is no such number print  - 1.

Input

The first line of the input contains an integer T (1 ≤ T ≤ 500), where T is the number of the test cases.

Each test case has one line that contains two integers n (1 ≤ n ≤ 106) number of digits, and s (1 ≤ s ≤ 9 × 106)
sum of digits.

Output

For each test case, print a single integer that represents the largest palindrome number that satisfies the mentioned rules, if there is no such number print  - 1.

Example

Input

2
2 2
4 26

Output

11
9449

题意：告诉你有一个n位数，这个数的各数位上的和为s，如果存在这样的回文数，则输出最大值，不存在输出-1。

分析：首先确定要把结果用字符串表示，然后判断不符合的条件，即（1）n为偶数s为奇数时 （2）9*n < s时 （3）n为奇数但不为1且s为1时。以上三种条件即输出-1。找到最大回文数的思路就是先把n位数全都赋值为9，然后从中间向两边递减，知道数位和等于s。然后对奇数要进行一次特殊处理，就是最中间的那一位，当s为奇数是，最中间那一位减到1就要停止，开始从两边减（因为之后的每次操作都是减2，只有现在的数位和为奇数才有可能使得总数位和符合条件）。

代码
b4d3
如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;
const int MX = 1e6 + 5;
const int mod = 1e9 + 7;
const int INF = 2e9 + 5;

char s[MX];

int main(){
int t;
scanf("%d", &t);
while(t--){
int n, ss;
scanf("%d%d", &n, &ss);
for(int i = 1; i <= n; i++){
s[i] = '9';
}
int sum = n * 9;
int mid = (1+n)/2;
if(sum < ss || (n % 2 == 0 && (ss & 1)) || (n & 1 && n != 1 && ss == 1)){
printf("-1\n");
continue;
}
if(n & 1){
while(sum > ss && s[mid] > '0'){
s[mid]--;
sum--;
if(s[mid] == '1' && (ss & 1)) break;
}
if(sum == ss){
for(int i = 1; i <= n; i++){
printf("%c", s[i]);
}
printf("\n");
continue;
}
else mid--;
while(sum > ss){
s[mid]--;
s[n - mid + 1]--;
sum -= 2;
if(s[mid] == '0')   mid--;
}
}
else{
while(sum > ss){
s[mid]--;
s[n - mid + 1]--;
sum -= 2;
if(s[mid] == '0')   mid--;
}
}
for(int i = 1; i <= n; i++){
printf("%c", s[i]);
}
printf("\n");
}
return 0;
}

 

A palindrome number is a number that can be read the same way from left to right and from right to left. For example: 961169, 111, 554455 are palindromes, while 856, 10180, 7226 are not.

Find the largest palindrome number that satisfies the following rules:

The number of its digits is equal to n, and must not contain leading zeros.
The sum of its digits is equal to s.
If there is no such number print  - 1.

Input

The first line of the input contains an integer T (1 ≤ T ≤ 500), where T is the number of the test cases.

Each test case has one line that contains two integers n (1 ≤ n ≤ 106) number of digits, and s (1 ≤ s ≤ 9 × 106)
sum of digits.

Output

For each test case, print a single integer that represents the largest palindrome number that satisfies the mentioned rules, if there is no such number print  - 1.

Example

Input

2
2 2
4 26

Output

11
9449