//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,sse4a,avx,avx2,popcnt,tune=native")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace std;
using namespace __gnu_pbds;
#define endl "\n"
#define fast ios_base::sync_with_stdio(false); cin.tie(nullptr);cout.tie(nullptr);
#define pb push_back
#define F first
#define S second
#define yes cout << "YES" << endl;
#define no cout<<"NO"<<endl;
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define MP(x,y) make_pair(x,y)
#define ll long long int
typedef tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef priority_queue<int> PQI;
const double Pii = 3.14159265358979323846;
const ll mod= 1e9+7;
const ll mod2= 998244353;
const ll inf= 1e16+7;
const int mxN=(2e5 + 5e4)*30+1;
const double eps=0.00000000000000001;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ll inv(ll i,ll mod) {if (i == 1) return 1; return (mod - ((mod / i) * inv(mod % i,mod)) % mod) % mod;}
/*
don't start implementing until you have a solid idea and you are 90% sure of it!
----
be always fully focused, sometimes we waste time thinking about nothing!
----
when we ask more questions we have a higher probability to find our way to the solution!
----
don't give up on a problem until you have questionned all the questions you can ask urself!
(trying greedy approaches , thinking about a trick ,about using a certain data structure
a certain algorithm,opting for a certain principle (e.g: pigeon principe) , solving the opposite problem,playing with examples in order to find
a pattern, thinking backwards, and most importantly, thinking about the box!
----
when u ask a question, try to prove it either it's correct or wrong,if it's wrong or it seems
it will take us nowhere =>try another approach.
----
don't get stuck at one approach!
----
apply teleportation, if we could do this? could we solve the next task?
----
implement very carefully!
----
RE : out_of_bounds, deleting an element from an empty data_structures, dividing by zero .
----
*/
/*
order_of_key (k) : Number of items strictly smaller than k .
find_by_order(k) : K-th element in a set (counting from zero).
*/
int k;
int to[mxN][2] ,firstTraverse[mxN][2];
int nodeCount=0;
int nextNode(){
return ++nodeCount;
}
// 0 is the root
void insert(int val,int ind){
int currNode=0;
for (int i = 30; i >= 0; i--) {
int b=(val>>i)&1;
if (to[currNode][b] == -1){
to[currNode][b]=nextNode();
}
currNode=to[currNode][b];
if(firstTraverse[currNode][b]==-1)firstTraverse[currNode][b]=ind;
}
}
int query(int val){
int currNode=0,mn=1e9;
for (int i = 30; i >= 0; i--) {
if(currNode==-1)return mn;
int b=!((val>>i)&1);
if (to[currNode][b] == -1){
if((1<<i)&k)return mn;
currNode=to[currNode][1^b];
}else{
if((1<<i)&k)currNode=to[currNode][b];
else{
mn=min(mn,firstTraverse[to[currNode][b]][b]);
currNode=to[currNode][1^b];
}
}
}
return mn;
}
void solve(){
int n;
cin>>n>>k;
k--;
memset(firstTraverse,-1,sizeof(firstTraverse));
memset(to,-1,sizeof(to));
insert(0,0);
int curr=0;
int l=2,r=1;
for(int i=1;i<=n;i++){
int x;
cin>>x;
curr^=x;
int mn=query(curr);
insert(curr,i);
assert(mn!=-1);
if(r-l<i-(mn+1)){
r=i;
l=mn+1;
}
}
cout<<l<<" "<<(r-l)+1<<endl;
}
int main(){
fast
// freopen("cases.in","r",stdin);
// freopen("output.txt","w",stdout);
// cout<<fixed<<setprecision(12);
int t=1;
// cin>>t;
while(t--)solve();
return 0;
}
/*
Keeeeeeeeeep goiiiiiing gonnaaaa do it now or after it's just a matter of time inchallah
*/
