CodeForces 622C Not Equal on a Segment（并查集）

题意：有n个数，然后m个询问，每次询问给你l,r,x，让你找到一个位置k，使得a[k]!=x,且l<=k<=r

思路：并查集，我们将a[i]=a[i-1]的合并在一起

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
int a[maxn];
int pre[maxn];
int Find(int x)
{
return x==pre[x]?x:pre[x]=Find(pre[x]);
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i = 1;i<=n;i++)
pre[i]=i;
for (int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
if (a[i]==a[i-1])
{
int u = Find(i);
int v = Find(i-1);
if (u!=v)
pre[u]=v;
}
}
for (int i = 1;i<=m;i++)
{
int flag = 0;
int l,r,x;
scanf("%d%d%d",&l,&r,&x);
int temp = r;
while (l<=temp)
{
if (a[temp]!=x)
{
printf("%d\n",temp);
flag=1;
break;
}
temp = Find(temp)-1;
}
if (flag==0)
printf("-1\n");
}
}

Description

You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Sample Input

Input

6 4

1 2 1 1 3 5

1 4 1

2 6 2

3 4 1

3 4 2

Output

2

6

-1

4