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 setBit(int n, int pos) { return n = n | (1 << pos); }
int resetBit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkBit(int n, int pos) { return (bool)(n & (1 << pos)); }
int n, m, A[1000000 + 10];
vector<int> pos[1000000 + 10];
int ok(int st, int ed, int x) {
  if (A[st] != x)
    return 1;
  else if (A[ed] != x)
    return 1;
  auto L = lower_bound(pos[x].begin(), pos[x].end(), st);
  auto R = lower_bound(pos[x].begin(), pos[x].end(), ed);
  int have = (R - L) + 1;
  if (have == (ed - st + 1)) {
    return -1;
  } else
    return 1;
}
int solve(int st, int ed, int x) {
  if (ok(st, ed, x) == -1) return -1;
  while (st < ed) {
    int mid = (st + ed) / 2;
    if (ok(st, mid, x) != -1)
      ed = mid;
    else
      st = mid + 1;
  }
  return st;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
    pos[A[i]].push_back(i);
  }
  int st, ed, x;
  while (m--) {
    scanf("%d %d %d", &st, &ed, &x);
    printf("%d\n", solve(st, ed, x));
  }
  return 0;
}