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], ans[maxn];
int c, d, e;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int r = 2;
  for (int i = 1; i <= n; i++) {
    while (a[i] == a[r] && r <= n) {
      r++;
    }
    ans[i] = r;
  }
  while (m--) {
    scanf("%d%d%d", &c, &d, &e);
    if (a[c] == e) {
      if (ans[c] <= d) {
        printf("%d\n", ans[c]);
        continue;
      } else {
        puts("-1");
      }
    } else {
      printf("%d\n", c);
      continue;
    }
  }
}