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;
const int N = 200008;
vector<pair<int, int> > t[N * 4];
int a[N];
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = vector<pair<int, int> >(1, make_pair(a[l], l));
  } else {
    int m = (l + r) >> 1;
    build(v + v, l, m);
    build(v + v + 1, m + 1, r);
    merge(t[v * 2].begin(), t[v * 2].end(), t[v * 2 + 1].begin(),
          t[v * 2 + 1].end(), back_inserter(t[v]));
  }
}
vector<pair<int, int> >::iterator it;
int get(int v, int l, int r, int L, int R, int x) {
  if (L > R) return -2;
  if (l == L && r == R) {
    it = lower_bound(t[v].begin(), t[v].end(), make_pair(x, -N));
    if (it != t[v].begin()) {
      it--;
      if (it->first != x) {
        return it->second;
      }
    }
    it = upper_bound(t[v].begin(), t[v].end(), make_pair(x, N));
    if (it != t[v].end() && it->first != x) {
      return it->second;
    }
    return -2;
  }
  int m = (l + r) >> 1;
  int ll = get(v + v, l, m, L, min(m, R), x);
  if (ll == -2) return get(v + v + 1, m + 1, r, max(m + 1, L), R, x);
  return ll;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  build(1, 0, n - 1);
  for (int i = 0; i < m; ++i) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    --l, --r;
    printf("%d\n", get(1, 0, n - 1, l, r, x) + 1);
  }
  return 0;
}