【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)。


代码如下:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#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

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