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 maxn = 200005;
const int inf = 1e9 + 5;
int n, q;
int niz[maxn];
int par[maxn], sz[maxn];
int findpar(int x) {
  if (x == par[x]) return x;
  return par[x] = findpar(par[x]);
}
void unite(int x, int y) {
  int a = findpar(x);
  int b = findpar(y);
  if (a == b) return;
  if (sz[a] > sz[b]) {
    par[a] = b;
    sz[b] += sz[a];
  } else {
    par[b] = a;
    sz[a] += sz[b];
  }
}
void init() {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cin >> n >> q;
  init();
  for (int i = 1; i <= n; i++) cin >> niz[i];
  for (int i = 2; i <= n; i++) {
    if (niz[i] == niz[i - 1]) unite(i, i - 1);
  }
  while (q--) {
    int l, r, x;
    cin >> l >> r >> x;
    if (niz[l] != x) {
      cout << l << endl;
      continue;
    }
    if (niz[r] != x) {
      cout << r << endl;
      continue;
    }
    if (findpar(l) == findpar(r)) {
      cout << -1 << endl;
      continue;
    }
    cout << findpar(l) + 1 << endl;
  }
  return 0;
}