Note: Except for the first subtask, the memory limit for this problem is 60MB.

Background

Zhengxu Sakiko is a famous competitive programmer. However, due to his poor memory, he often forgets the algorithms he has learned.

This is why he often reminisces about the past.

Please pay attention to the special memory limit for this problem.

Given $m$ triples $(l, r, v)$.

There are $q$ operations. Each operation provides $x$ and $y$, representing a query for the maximum value of $v$ among all triples such that $l \le x \le r$ and $v \le y$.

If no such triple exists, output $0$.

This problem is forced online.

Input

The first line contains two integers $m$ and $q$.

The next $m$ lines each contain three integers representing a triple $(l, r, v)$.

The next $q$ lines each contain two integers $p$ and $q$. The current query parameters are $x = p \oplus ans$ and $y = q \oplus ans$, where $\oplus$ denotes the bitwise XOR operation, and $ans$ is the bitwise AND of the previous query's answer and $2^{21}-1$. The initial $ans$ is $0$.

Output

For each query, output the answer on a new line.

Examples

Input 1

5 5
3 5 3
4 5 10
3 4 2
1 4 13
1 1 6
1 6
7 11
9 14
2 1
5 14

Output 1

6
13
3
0
10

Input 2

15 15
3 9 9
3 5 10
8 15 10
4 6 4
8 13 6
3 7 0
11 11 11
8 10 11
3 3 14
3 15 13
4 15 9
6 11 7
4 5 12
6 12 10
10 10 14
9 5
1 5
8 11
14 9
8 14
3 11
7 1
4 3
3 1
12 15
9 4
13 15
0 3
13 12
6 13

Output 2

0
0
11
0
13
0
0
0
0
13
9
4
4
7
0

Subtasks

For all data: $1 \le m, q \le 10^6$, $1 \le l \le r \le 2 \times 10^6$, $0 \le v < 2^{31}$, $1 \le x \le 2 \times 10^6$, $0 \le y < 2^{31}$.

Note

Additionally, here is some code written by Little Z for finding the $k$-th smallest value in a range, which might be helpful to you.

#include <bits/stdc++.h>
using namespace std;
constexpr int Spp{1<<20},S2{1<<20};
char buf[Spp],*p1,*p2,buf2[S2],*l2=buf2,_st[22];
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,Spp,stdin),p1==p2)?EOF:*p1++)
#define putchar(c) (l2==buf2+S2&&(fwrite(buf2,1,S2,stdout),l2=buf2),*l2++=(c))
template <typename T>
void read(T &x) {
    char c;int f{1};
    do x=(c=getchar())^48;
    while (!isdigit(c)&&c!='-');
    if (x==29) f=-1,x=0;
    while (isdigit(c=getchar()))
        x=(x*10)+(c^48);
    x*=f;
}
template <typename T,typename ...Args>
void read(T& x,Args&... args) {read(x);read(args...);}
template<typename T>
void write(T x,char c='\n') {
    if (x<0) putchar('-'),x*=-1;
    int tp=0;
    do _st[++tp]=x%10; while (x/=10);
    while (tp) putchar(_st[tp--]+'0');
    putchar(c);
}
struct OI{~OI(){fwrite(buf2,1,l2-buf2,stdout);}}oi;
using LL=unsigned long long;
// wavelet
struct bits {
    vector<LL> bl;
    vector<int> c;
    void resize(int num) {
        bl.resize(((num+1)>>6)+1);
        c.resize(bl.size());
    }
    void set(int i,LL val) {bl[i>>6]|=(val<<(i&63));}
    void build() {
        for (int i=1;i<bl.size();++i)
            c[i]=c[i-1]+popcount(bl[i-1]);
    }
    int rk1(int i) {return c[i>>6]+popcount(bl[i>>6]&((1uLL<<(i&63))-1));}
    int rk0(int i) {return i-rk1(i);}
};
struct wavelet {
    int h;
    vector<bits> B;
    vector<int> pos;
    void init(vector<int>& v) {
        h=__lg(*max_element(v.begin(),v.end()))+1;
        B.resize(h);pos.resize(h);
        for (int i=0;i<h;++i) {
            B[i].resize(v.size());
            for (int j=0;j<v.size();++j)
                B[i].set(j,(v[j]>>(h-i-1)&1));
            B[i].build();
            pos[i]=stable_partition(v.begin(),v.end(),[&](int c){
                return !(c>>(h-i-1)&1);
            })-v.begin();
        }
    }
    int qry(int i,int j,int k,int l,int r,int x) {
        if (i==j) return 0;
        if (r==l+1) return l;
        int mid=l+r>>1;
        int L{B[x].rk0(i)},R{B[x].rk0(j)};
        if (R-L>=k) return qry(L,R,k,l,mid,x+1);
        else return qry(pos[x]+B[x].rk1(i),pos[x]+B[x].rk1(j),k-(R-L),mid,r,x+1);
    }
} tr;
int main() {
    int n,m;read(n,m);
    vector<int> a(n);
    for (int i=0;i<n;++i) read(a[i]);
    tr.init(a);
    while (m--) {
        int l,r,k;read(l,r,k);
        write(tr.qry(l-1,r,k,0,1<<tr.h,0));
    }
    return 0;
}