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 ax[200010];
int bn[200010];
int main(void) {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ax[i]);
  }
  int gg = n - 1;
  bn[gg] = gg;
  for (int i = n - 2; i >= 0; i--) {
    if (ax[i] != ax[gg]) {
      bn[i] = gg;
      gg = i;
    } else {
      bn[i] = bn[gg];
      gg--;
    }
  }
  while (m--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    if (bn[l - 1] <= r - 1) {
      if (ax[l - 1] == x && (bn[l - 1] != n - 1 || ax[bn[l - 1]] != x))
        printf("%d\n", bn[l - 1] + 1);
      else if (ax[l - 1] == x) {
        printf("-1\n");
      } else
        printf("%d\n", l);
    } else {
      if (ax[l - 1] == x)
        printf("-1\n");
      else
        printf("%d\n", l);
    }
  }
}