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 = 3e5 + 10;
int a[maxn];
int b[maxn];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int x1, cnt = 0;
  scanf("%d", &a[0]);
  b[0] = 1;
  for (int i = 1; i < n; i++) {
    scanf("%d", &x1);
    if (x1 == a[cnt])
      b[cnt]++;
    else {
      cnt++;
      b[cnt] = b[cnt - 1] + 1;
      a[cnt] = x1;
    }
  }
  cnt++;
  while (m--) {
    int l, r, x, need;
    scanf("%d %d %d", &l, &r, &x);
    int k = lower_bound(b, b + cnt, l) - b;
    if (b[k] >= r) {
      if (a[k] == x)
        printf("-1\n");
      else
        printf("%d\n", r);
    } else {
      if (a[k] == x)
        printf("%d\n", b[k] + 1);
      else
        printf("%d\n", b[k]);
    }
  }
}