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 = 2e5 + 1;
vector<pair<int, int> > p[500], a(N);
vector<int> q[500];
int main() {
  int n, m, root, rq;
  scanf("%d%d", &n, &m);
  root = sqrt(n);
  rq = (n + root - 1) / root;
  for (int i = 0, x; i < n; i++) {
    scanf("%d", &x);
    a[i].first = x;
    a[i].second = i + 1;
    p[i / root].push_back(make_pair(x, i + 1));
    q[i / root].push_back(x);
  }
  for (int i = 0; i < rq; i++) {
    sort(q[i].begin(), q[i].end());
    q[i].resize(unique(q[i].begin(), q[i].end()) - q[i].begin());
  }
  for (int i = 0; i < m; i++) {
    int l, r, z, ans = -1;
    scanf("%d%d%d", &l, &r, &z);
    l--;
    r--;
    for (int i2 = l; i2 <= r; i2++) {
      if (i2 % root == 0 && i2 + root - 1 <= r) {
        i2 += root - 1;
        if (q[i2 / root].size() == 1 && q[i2 / root][0] == z) continue;
        for (int i3 = 0; i3 < p[i2 / root].size(); i3++)
          if (p[i2 / root][i3].first != z) {
            ans = p[i2 / root][i3].second;
            break;
          }
      } else {
        if (a[i2].first != z) {
          ans = a[i2].second;
          break;
        }
      }
      if (ans != -1) {
        break;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}