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 double PI = 2.0 * acos(0.0);
const double EX = 2.7182818284;
const int MOD = 1e9 + 7;
const int oo = 2e9 + 1e8;
long long gcd(long long x, long long y) { return !y ? x : gcd(y, x % y); }
const int mx = 1e6 + 10;
int n, m, l, r, x, ans;
pair<int, int> idx[mx];
bool vis[mx];
int main() {
  scanf("%d%d", &n, &m);
  vector<int> in(n + 1), dp(n + 1);
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &in[i]);
    if (in[i] != in[i - 1]) dp[i] = 1;
  }
  in[0] = -1;
  idx[1] = make_pair(0, 1);
  for (int i = 2; i < n + 1; i++) {
    if (dp[i])
      idx[i] = make_pair(i - 1, i);
    else
      idx[i] = make_pair(idx[i - 1].first, i);
  }
  while (m-- && scanf("%d%d%d", &l, &r, &x)) {
    if (l == r) {
      if (in[idx[r].second] != x)
        ans = idx[r].second;
      else
        ans = -1;
    } else {
      if (in[idx[r].second] != x)
        ans = idx[r].second;
      else if (idx[r].first >= l && in[idx[r].first] != x)
        ans = idx[r].first;
      else
        ans = -1;
    }
    printf("%d\n", ans);
  }
  return 0;
}