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;
inline void splay(int &v) {
  v = 0;
  char c = 0;
  int p = 1;
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    v = (v << 3) + (v << 1) + c - '0';
    c = getchar();
  }
  v *= p;
}
int a[200010], t[200010];
int main() {
  int n, m;
  splay(n);
  splay(m);
  for (int i = 1; i <= n; i++) splay(a[i]);
  t[n] = n + 1;
  for (int i = n - 1; i >= 1; i--) {
    if (a[i] == a[i + 1])
      t[i] = t[i + 1];
    else
      t[i] = i + 1;
  }
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    splay(l), splay(r), splay(x);
    if (a[l] != x)
      printf("%d\n", l);
    else {
      if (t[l] <= r)
        printf("%d\n", t[l]);
      else
        printf("-1\n");
    }
  }
  return (0);
}