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 int MAXN = 2 * int(1e5) + 5;
const int MAXL = 20;
const int DEFAULT = -2;
int n, m;
int vtx[MAXN];
pair<int, int> a[MAXL][MAXN];
class Solution {
 public:
  int get(pair<int, int> c, int other) {
    int i1 = c.first, i2 = c.second;
    int v1 = i1 >= 0 ? vtx[i1] : 0;
    int v2 = i2 >= 0 ? vtx[i2] : 0;
    if (v1 != 0 && v1 != other) {
      return i1;
    } else if (v2 != 0 && v2 != other) {
      return i2;
    }
    return DEFAULT;
  }
  int get(pair<int, int> c1, pair<int, int> c2, int other) {
    int ans = get(c1, other);
    if (ans == DEFAULT) {
      ans = get(c2, other);
    }
    return ans;
  }
  void build() {
    for (int i = 0, lets_stop_here = (int)(n); i < lets_stop_here; i++) {
      a[0][i].first = i;
      a[0][i].second = DEFAULT;
    }
    for (int i = 1, lets_stop_here = (int)(MAXL - 1); i <= lets_stop_here;
         i++) {
      for (int j = 0, lets_stop_here = (int)(n - (1 << i) + 1);
           j < lets_stop_here; j++) {
        int d = 1 << (i - 1);
        pair<int, int> f = a[i - 1][j];
        pair<int, int> s = a[i - 1][j + d];
        int cx = get(f, s, 0);
        int cy = get(f, s, cx == DEFAULT ? 0 : vtx[cx]);
        a[i][j].first = cx;
        a[i][j].second = cy;
      }
    }
  }
  void solve(std::istream& in, std::ostream& out) {
    in >> n >> m;
    for (int i = 0, lets_stop_here = (int)(n); i < lets_stop_here; i++) {
      in >> vtx[i];
    }
    build();
    for (int i = 0, lets_stop_here = (int)(m); i < lets_stop_here; i++) {
      int l, r, vx, ans = DEFAULT;
      in >> l >> r >> vx;
      l--, r--;
      for (int j = MAXL - 1, lets_stop_here = (int)(0); j >= lets_stop_here;
           j--) {
        int d = (1 << j);
        if (l + d - 1 <= r) {
          ans = get(a[j][l], vx);
          l += d;
        }
        if (ans > -1) {
          break;
        }
      }
      out << ans + 1 << ("\n");
    }
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}