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>
int a[200010];
int b[200010];
using namespace std;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int i, j;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = 1;
  }
  for (i = 1; i <= n; i++) {
    int l = i;
    while (i < n && a[i] == a[i + 1]) i++;
    int len = i - l + 1;
    if (len == 1) continue;
    for (j = l; j <= i; j++) {
      b[j] = len;
      len--;
    }
  }
  for (i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (a[x] != z) {
      printf("%d\n", x);
      continue;
    }
    if (b[x] >= (y - x + 1)) {
      printf("-1\n", x);
      continue;
    }
    printf("%d\n", x + b[x]);
  }
  return 0;
}