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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
int pre[200055], a[200055];
void fmain() {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  pre[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == a[i - 1])
      pre[i] = pre[i - 1] + 1;
    else
      pre[i] = 1;
  }
  for (int l, r, x, i = 0, ans; i < m; i++) {
    scanf("%d%d%d", &l, &r, &x);
    ans = -1;
    if (a[r] != x)
      ans = r;
    else {
      if (r - pre[r] >= l) ans = r - pre[r];
    }
    printf("%d\n", ans);
  }
}
int main() {
  fmain();
  return 0;
}