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;
int dX[] = {-1, 0, 1, 0, -1, 1, 1, -1};
int dY[] = {0, 1, 0, -1, 1, 1, -1, -1};
inline void Refresh() {}
int n, m;
int a[2 * 100007];
int lft[2 * 100007];
void preprocess() {
  memset(lft, -1, sizeof lft);
  for (int i = 2; i <= n; ++i) {
    lft[i] = ((a[i] != a[i - 1]) ? (i - 1) : lft[i - 1]);
  }
  return;
}
int main() {
  int l, r, x;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  preprocess();
  for (int i = 0; i < m; ++i) {
    scanf("%d %d %d", &l, &r, &x);
    if (a[r] != x)
      printf("%d\n", r);
    else {
      if (lft[r] >= l)
        printf("%d\n", lft[r]);
      else
        printf("-1\n");
    }
  }
  return 0;
}