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;
double pi = 2 * acos(0.0);
const int INF = 1e9 + 7;
long long gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long int a, long long int b) { return (a / gcd(a, b)) * b; }
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
void WetPitch() {
  long long int n, q;
  cin >> n >> q;
  vector<long long int> v(n + 1);
  long long int z[n + 1];
  for (int i = 1; i < n + 1; i++) {
    cin >> v[i];
  }
  z[n] = -1;
  for (int i = n - 1; i >= 1; i--) {
    if (v[i] != v[i + 1]) {
      z[i] = i + 1;
    } else {
      z[i] = z[i + 1];
    }
  }
  for (int i = 0; i < q; i++) {
    long long int l, r, x;
    cin >> l >> r >> x;
    if (v[l] != x) {
      cout << l << endl;
    } else {
      if (z[l] <= r) {
        cout << z[l] << endl;
      } else {
        cout << -1 << endl;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tst;
  tst = 1;
  while (tst--) {
    WetPitch();
  }
  return 0;
}