In this HackerEarth Boring Not boring problem solution You are given two integers n(1 <= n <= 10^5) and m(1 <= m <= 10^5). Initially, you have an array a of n zeros, and m applied queries on it. Query is given as l r x, where you apply ai = ai xor x (xor denotes the operation bitwise XOR) for every i between l and r. But this problem seems boring, so Almas modify some queries interval. By modifying, he applied that operation on a subset of positions on a segment(maybe an empty subset). Now he asks you how many different arrays he could have after applying queries. Since the answer can be large, output by modulo 1000000007 (10^9 + 7).

## HackerEarth Boring Not boring problem solution.

```# include <bits/stdc++.h>

# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define _USE_MATH_DEFINES_
#define ll long long
#define ld long double
#define Accepted 0
#define pb push_back
#define mp make_pair
#define sz(x) (int)(x.size())
#define every(x) x.begin(),x.end()
#define F first
#define S second
#define lb lower_bound
#define ub upper_bound
#define For(i,x,y)  for (ll i = x; i <= y; i ++)
#define FOr(i,x,y)  for (ll i = x; i >= y; i --)
#define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)

void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void setIO(string s = "") {
if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

const double eps = 0.000001;
const ld pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const ll MOD = 1e18 + 9;
const ll INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 1e5+5;
const int M = 22;
const int pri = 997;
const int Magic = 2101;

const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};

int rnd (int l, int r) {
return uniform_int_distribution<int> (l, r)(gen);
}

int n, m;

vector < int > g[N*4];

struct GG {
//gauss
vector < int > basis;
for(auto &it:basis) if((x^it)<x) x^=it;
for(auto &it:basis) if((it^x)<it) it^=x;
if(x) basis.pb(x);
}

int get() {
int ans = 0;
for (auto x:basis) ans ^= x;
return ans;
}

void clear() {
basis.clear();
}
};

void add(int l, int r, int x, int v=1, int tl = 1, int tr =n) {
if(tl>r || l > tr) return;
if(tl >= l && tr <= r) {
g[v].pb(x);
return;
}

int tm = (tl+tr)>>1;
add(l, r, x, v<<1, tl, tm);
add(l, r, x, v<<1|1, tm+1, tr);
}

int pw[33];

int calc(GG c, int v, int l, int r) {
for (auto x : g[v])

if(l == r) {
return pw[c.basis.size()];
}

int md = (l+r)>>1;
return calc(c, v<<1, l, md)
* (ll)calc(c, v<<1|1, md+1, r) % mod;
}

void solve() {
cin >> n >> m;

for (int i = 1, l, r, x; i <= m; ++i) {
cin >> l >> r >> x;
}

GG c;
cout << calc(c, 1, 1, n) << '\n';
}

int main () {
SpeedForce;
pw[0] = 1;

for (int i = 1; i < 33; ++i) {
pw[i] = (pw[i-1] * 2) % mod;
}

int T = 1;
//cin >> T;
while(T--) solve();

return Accepted;
}```

### Second solution

```#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int MAX_N = 1e5 + 14, LG = 30, MOD = 1000000007;

struct Gauss {
int a[LG];

Gauss() {
memset(a, 0, sizeof a);
}

for (int i = LG - 1; i >= 0; i--)
if (x >> i & 1)
if (a[i])
x ^= a[i];
else {
a[i] = x;
break;
}
}

void operator+=(vector<int> &o) {
for (int i : o)
if (i)
}

int cnt() {
return 1 << LG - count(a, a + LG, 0);
}
};

vector<int> node[1 << 18];
int n;

void add(int s, int e, int x, int l = 0, int r = n, int id = 1) {
if (s <= l && r <= e) {
node[id].push_back(x);
return;
}
if (e <= l || r <= s)
return;
int mid = (l + r) / 2;
add(s, e, x, l, mid, id * 2);
add(s, e, x, mid, r, id * 2 + 1);
}

int get_ans(int l = 0, int r = n, int id = 1, Gauss carry = Gauss()) {
carry += node[id];
if (l + 1 == r)
return carry.cnt();
int mid = (l + r) / 2;
return (ll) get_ans(l, mid, id * 2, carry) * get_ans(mid, r, id * 2 + 1, carry) % MOD;
}

int main() {
ios::sync_with_stdio(0), cin.tie(0);
int m;
cin >> n >> m;
for (int i = 0; i < m; ++i) {
int l, r, x;
cin >> l >> r >> x;