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 = 2e5 + 5;
int p[maxn], a[maxn];
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      if (i == 1)
        p[i] = -1;
      else if (a[i] == a[i - 1])
        p[i] = p[i - 1];
      else
        p[i] = i - 1;
    }
    while (q--) {
      int l, r, x, ans;
      scanf("%d%d%d", &l, &r, &x);
      if (a[r] != x)
        ans = r;
      else if (p[r] >= l)
        ans = p[r];
      else
        ans = -1;
      printf("%d\n", ans);
    }
  }
  return 0;
}