#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    ifstream fin("note2.in");
    ofstream fout("note2.out");
    
    int N;
    ll M;
    fin >> N >> M;
    vector<ll> a(N);
    for (int i = 0; i < N; i++){
        fin >> a[i];
    }
    
    // Special-case: if M == 0, every subarray qualifies.
    if(M == 0){
        ll total = (ll)N * (N + 1) / 2;
        fout << total;
        return 0;
    }
    
    // dp will hold pairs (value, count) for distinct cumulative AND values
    // of subarrays ending at the previous index.
    vector<pair<ll, ll>> dp;
    ll answer = 0;
    
    for (int i = 0; i < N; i++){
        vector<pair<ll, ll>> newdp;
        // The subarray that starts and ends at i
        newdp.push_back({a[i], 1});
        
        // Extend each subarray ending at i-1 by including a[i].
        for (auto &p : dp) {
            ll new_val = p.first & a[i];
            // If the last pair in newdp has the same value, merge the count.
            if (!newdp.empty() && newdp.back().first == new_val)
                newdp.back().second += p.second;
            else
                newdp.push_back({new_val, p.second});
        }
        
        // Count subarrays ending at i whose AND >= M.
        for (auto &p : newdp) {
            if(p.first >= M)
                answer += p.second;
        }
        
        dp = move(newdp);
    }
    
    fout << answer;
    return 0;
}