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;
inline int in() {
  int res = 0;
  char c;
  int f = 1;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
const int N = 200010;
int a[N];
int dp[N];
int main() {
  int n = in(), m = in();
  for (int i = 1; i <= n; i++) {
    a[i] = in();
  }
  for (int i = n; i >= 1; i--) {
    if (a[i] == a[i + 1]) dp[i] = dp[i + 1] + 1;
  }
  while (m--) {
    int l = in(), r = in(), x = in(), ans = -1;
    for (int i = l; i <= r; i += dp[i] + 1) {
      if (a[i] != x) {
        ans = i;
        break;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}