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 mod = 1e9 + 7;
const int N = 1e6 + 10;
const int inf = 2e9;
const double eps = 1e-7;
const long long INF = (long long)1e18;
vector<int> v;
int a[N];
int nxt[N];
int n, m, l, r, x;
int main() {
  int lst = 0;
  scanf("%d%d", &(n), &(m));
  for (int i = (1); i <= (n); ++i) scanf("%d", &(a[i]));
  nxt[n] = n + 1;
  for (int i = n - 1; i > 0; i--) {
    if (a[i] == a[i + 1])
      nxt[i] = nxt[i + 1];
    else
      nxt[i] = i + 1;
  }
  while (m--) {
    scanf("%d%d%d", &l, &r, &x);
    if (a[l] != x)
      cout << l << "\n";
    else if (nxt[l] <= r)
      cout << nxt[l] << "\n";
    else
      puts("-1");
  }
  return 0;
}