Problem Description

For a positive integer?n, let's denote function?f(n,m)?as the?m-th smallest integer?x?that?x>n?and?gcd(x,n)=1. For example,?f(5,1)=6?and?f(5,5)=11.

You are given the value of?m?and?(f(n,m)?n)⊕n, where ``⊕'' denotes the bitwise XOR operation. Please write a program to find the smallest positive integer?nthat?(f(n,m)?n)⊕n=k, or determine it is impossible.

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, there are two integers k,m(1≤k≤1018,1≤m≤100).

Output

For each test case, print a single line containing an integer, denoting the smallest n. If there is no solution, output ``-1'' instead.

Sample Input

2
3 5
6 100

Sample Output

5
-1

題意：t 組數據，函數 f(n,m) 表示比 n 大的第 m 小的與 m 互質的數，現在每組給出 k、m 兩個數，對于式子 (f(n,m)-n)⊕n=k，求最小的 n，如果沒有答案，輸出 -1

思路：

對于 (f(n,m)-n)⊕n=k

設 d=f(n,m)-n

那么根據異或的性質，有：d=n⊕k

故而：n=d⊕k

根據質數分布密度定理：素數的分布越來越稀疏，當 1E18 內的任意兩個素數的差不會很大（不會超過 300），因此可以直接枚舉 d 的值來尋找答案

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<unordered_map> #include<bitset> #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long #define Pair pair<int,int> LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; } LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;} LL quickPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;} LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); } LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); } LL LCM(LL x,LL y){ return x/GCD(x,y)*y; } const double EPS = 1E-10; const int MOD = 998244353; const int N = 1000+5; const int dx[] = {-1,1,0,0,1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;LL cal(LL n, int m) {if (n < 1) return 0;else{LL i=n+1;while(true){if (GCD(n, i) == 1) {m--;if (!m)return i - n;}i++;}} } int main() {int t;scanf("%d", &t);while (t--) {LL k;int m;scanf("%lld%d", &k, &m);LL res = -1;for (int d = 1; d <= 1000; d++) {if (cal(k ^ d, m) == d) {if (res == -1)res = k ^ d;else if (res > (k ^ d))res = k ^ d;}}printf("%lld\n", res);}return 0; }

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