#include <bits/stdc++.h> using namespace std; typedef signed long long ll; #define _P(...) (void)printf(__VA_ARGS__) #define FOR(x,to) for(x=0;x<(to);x++) #define FORR(x,arr) for(auto& x:arr) #define FORR2(x,y,arr) for(auto& [x,y]:arr) #define ALL(a) (a.begin()),(a.end()) #define ZERO(a) memset(a,0,sizeof(a)) #define MINUS(a) memset(a,0xff,sizeof(a)) template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;} template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;} //------------------------------------------------------- ll N,K; ll dp[62][65][2]; const ll mo=998244353; void solve() { int i,j,k,l,r,x,y; string s; cin>>N>>K; dp[0][0][0]=1; FOR(i,61) { int n=(N>>i)%2; int k=(K>>i)%2; FOR(j,62) { FOR(x,2) { // not take (dp[i+1][j+(k^x)][k&x]+=dp[i][j][x])%=mo; if(n) { (dp[i+1][j+(k^x^1)][((k+x+1)>>1)]+=dp[i][j][x])%=mo; } } } } ll ret=0; FOR(j,62) (ret+=dp[61][j][0]%mo*((1LL<<j)%mo))%=mo; cout<<ret<<endl; } int main(int argc,char** argv){ string s;int i; if(argc==1) ios::sync_with_stdio(false), cin.tie(0); FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin); cout.tie(0); solve(); return 0; }