Problem: 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.

Examples

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[2 * 100000 + 100];
int p[2 * 100000 + 100];
int n, m;
int li, ri, xi;
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == a[i - 1])
        p[i] = p[i - 1];
      else
        p[i] = i - 1;
    }
    while (m--) {
      scanf("%d%d%d", &li, &ri, &xi);
      if (a[ri] == xi)
        if (p[ri] >= li)
          printf("%d\n", p[ri]);
        else
          printf("-1\n");
      else
        printf("%d\n", ri);
    }
  }
  return 0;
}