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;
int Scan() {
  int res = 0, ch, flag = 0;
  if ((ch = getchar()) == '-')
    flag = 1;
  else if (ch >= '0' && ch <= '9')
    res = ch - '0';
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - '0';
  return flag ? -res : res;
}
void Out(int a) {
  if (a > 9) Out(a / 10);
  putchar(a % 10 + '0');
  putchar('\n');
}
int dp[200005];
int a[200005];
int n, m;
int main() {
  n = Scan();
  m = Scan();
  for (int i = 1; i <= n; i++) a[i] = Scan();
  dp[1] = -1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[i - 1])
      dp[i] = dp[i - 1];
    else
      dp[i] = i - 1;
  }
  for (int i = 0; i < m; i++) {
    int l, r, x;
    l = Scan(), r = Scan(), x = Scan();
    if (x != a[r])
      printf("%d\n", r);
    else if (dp[r] >= l)
      printf("%d\n", dp[r]);
    else
      printf("-1\n");
  }
  return (0);
}