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 = 1e6 + 10;
int n, m;
int a[maxn];
int d[maxn];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int A = a[1];
  int LA = 1;
  int B = 0;
  int LB = 0;
  d[1] = -1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == A) {
      d[i] = LB;
      LA = i;
    } else if (a[i] == B || B == 0) {
      B = a[i];
      LB = i;
      d[i] = LA;
    } else {
      if (LA < LB) {
        d[i] = i - 1;
        A = a[i];
        LA = i;
      } else {
        d[i] = i - 1;
        B = a[i];
        LB = i;
      }
    }
  }
  int l, r, x;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &l, &r, &x);
    if (l == r) {
      if (a[l] == x)
        printf("-1\n");
      else
        printf("%d\n", l);
    } else if (d[r] < l && a[r] == x)
      printf("-1\n");
    else {
      if (a[r] != x)
        printf("%d\n", r);
      else {
        printf("%d\n", d[r]);
      }
    }
  }
  return 0;
}