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;
  scanf("%d %d", &n, &m);
  vector<int> v(n + 3);
  vector<vector<int> > freq(1000050);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    freq[v[i]].push_back(i);
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    if (upper_bound(freq[x].begin(), freq[x].end(), r) -
            lower_bound(freq[x].begin(), freq[x].end(), l) ==
        r - l + 1) {
      printf("-1\n");
      continue;
    }
    int lo = l, hi = r;
    while (hi - lo > 3) {
      int mid = (lo + hi) / 2;
      if (upper_bound(freq[x].begin(), freq[x].end(), mid) -
              lower_bound(freq[x].begin(), freq[x].end(), l) ==
          mid - l + 1) {
        lo = mid;
      } else {
        hi = mid;
      }
    }
    int k = 0;
    for (int mid = lo; mid <= hi; mid++) {
      if (v[mid] != x) {
        printf("%d\n", mid);
        k = 1;
        break;
      }
    }
    if (k == 0) {
      printf("%d\n", hi + 1);
    }
  }
  return 0;
}