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 main() {
  int n, m, t, l, r, x;
  vector<int> a;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &t);
    a.push_back(t);
  }
  vector<int> b;
  b.resize(n, -1);
  for (int i = 0; i < n; i++) {
    if (i == 0 || a[i] != a[i - 1]) {
      int j = i + 1;
      b[i] = -1;
      while (j < n && a[i] == a[j]) {
        j++;
      }
      if (j < n) {
        b[i] = j;
      }
    } else {
      b[i] = b[i - 1];
    }
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &l, &r, &x);
    l--;
    r--;
    if (a[l] != x) {
      printf("%d\n", l + 1);
    } else {
      if (b[l] <= r && b[l] != -1) {
        printf("%d\n", b[l] + 1);
      } else {
        printf("%d\n", -1);
      }
    }
  }
  return 0;
}